example problems - users.ugent.beusers.ugent.be/~vstorme/files/math/calculus-1.pdf · what is the...
TRANSCRIPT
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EXAMPLE PROBLEMS(done in the text)
• A lake is stocked with 1000 fish. It is found that N(t), the number of fish after t years, increases so that its rate of
increase is governed by the equation: dNdt = k N (10,000 - N) (called the "Logistic Equation"). Show that the
rate of change, dNdt , is a maximum when the population is 5,000 fish.
• The cost of manufacturing an item is $100 even if no items are manufactured ... and the cost decreases with eachitem; for x items the cost per item is 100 - .1x (i.e. the cost decreases by $0.1 for each item). Graph C(x), thecost of producing x items.
• An orchard contains 240 apple trees, each tree producing 30 bushels of apples. For each additional tree planted,the yield per tree decreases by 1/12 bushel (due to overcrowding). Sketch N(x), the total apple production as afunction of x, the number of additional trees planted.
• A conical drinking cup is formed from a circular piece of paper by removing a sector and joining the edges. Ifthe radius of the piece of paper is 10 cm., what should be the angle q so as to yield a cup of maximum volume?
• A man can run 10 times faster than he can swim. He begins in the water at a point P, swims to shore, then runs toQ. Describe his path so the total time is a minimum.
• A certain amount of money is left in the bank to accumulate interest (compounded at i% per year). If you want todouble your money in n years, what should the interest rate be?
• In drilling a well, the cost per metre depends upon the type of sand, gravel or rock which must be excavated.Suppose the cost is C(x) dollars/metre at a depth x metres. (As x changes, the type of material changes, hencethe cost changes.) Express, as a definite integral., the cost in digging a well of H metres.
• The cost of manufacturing an item depends upon the number of items manufactured. For the first few items thecost is high and the profit low, but as we produce more items the cost decreases hence the profits (when we sell
the items) increase. Suppose the profit for the nth item is p(n) dollars per item. Express, as a definite integral, theprofit in producing (and selling) K items.
• You invest $10,000 in a mutual fund, then, 5 months later you put an additional $15,000 into the fund, then, 3months later put in an additional $5,000. At the end of a year, your investments (totalling $30,000) have grownto $31,470. What is the annual rate of return from this mutual fund?
ASSORTED PROBLEMS(which you'll be able to solve by the end of this course)
1. Evaluate:
(a)limx->2
|x2-8| - |2+x|x-2
(b) f '(2) i f f(x) = (x-2) sin πx using the limit definition of derivative!
(c) f '(- π3 ) if f(t) = | sin t | (you needn't use the definition ... unless you want to)
2. Compute the area bounded by the curves y = x2 + 1x + 1 and x + 3y = 7. (Include a sketch, and since this is a
complicated evaluation with lots of room for errors, check for reasonableness.)
3. Evaluate ⌡⌠ x - 3
x2+2x+1 dx
4. Use Newton's Method to find approximations to the roots of the following equations, correct to five decimalplaces. In each case, make a reasonable plot of the function in order to obtain an the initial "guess", x1.
(a) x3 - x2 + x - 22 = 0 (b) x ln x = 65. Use l'Hopital's Rule to evaluate the following limits:
(a)limx->0
1 - cos2x
x2 (b)
limx->0
ex - 1 - x - x2
x2
6: Express each of the following in terms of one or more definite integrals using both HORIZONTAL and
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VERTICAL rectangles. DO NOT EVALUATE but include a sketch:(a) The area bounded by y = 0, y = x, y = ln x, and y = 1.
(b) The volume when the region described by 0 ≤ y ≤ 1 - x2, x ≥ 0, is revolved about the y-axis.7. Evaluate each of the following limits (or explain why the limit does not exist):
(a)limx->0
|2x-1| - |2x+1|x (b)
limx->-∞
( )x2+2x - x2+x
8. Evaluate:
(a) cos
arctan(-
12) (b)
ddx
arcsin x + 1-x2
x ... and simplify your answer
9. Calculate:
(a) the average value of f(x) = 1
4 + x2 on 0 ≤ x ≤ 2 (b) ⌡⌠
-1
1
10x dx
10. Evaluate: (a)lim
L->1-
⌡⌠
0
Ldx
1-x2 (b)
ddx
⌡⌠
-x2
x2
t sin t dt
11. The function f(x) = 4 x3
x2 + 1 , with domain x≥ 0, has an inverse g(x).
(a) Test f(x) to verify that it does have an inverse when x ≥ 0(b) Calculate g(2) (c) Calculate g'(2)
12. The numbers π and e are both near "3". Which of eπ or πe is larger?
13. Determine the shortest distance from the origin to the curve y = (3 - x2)/2. Include a sketch!
14. Find dydx if : (a) y =
ln (ln x)ln x (b) y = x cos (ln x)
15. Calculate the areas described (and include a sketch):
(a) below y = x and above y = x2
(b) enclosed by y = 1
x3 , y = 0, x = -4 and x = -2.
16. Evaluatedydx if y = ⌡⌠
-t2
t2
sin x3 dx
17. (a) If y = ax, determine dydx by first taking the ln of each side.
(b) If y = xx, determine dydx by first taking the ln of each side.
18. Determine an approximate value to 641/3 by using a linear approximation.
19. Evaluate⌡⌠ dx
ex + e-x
20. Sketch the graph of f(x) = x2-1
x2+x showing where f(x) is increasing, decreasing, asymptotes and all critical points.
21. Find the area bounded by y = 4xπ and y = tan x, between x = 0 and the first intersection of these two curves to
the right of x = 0. (Include a reasonable sketch.)22. Calculate the area for each of the following regions (bounded by polar curves):
(a) inside r = sin q(b) inside the cardioid r = 1 + sin q and outside the circle r = 1(c) inside the smaller loop of the limaçon r = 1 - 2 sin q
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(d) inside the lemniscate r2 = cos 2q23. Aluminum pop cans to hold 300 mL are made in the shape of right circular cylinders. Find the dimensions which
minimize the amount of aluminum used.24. A rectangular field next to an ocean is to be fenced on three sides with 1000 m of fencing. (The fourth side,
being the shoreline, is not fenced.) Determine the dimensions of the field so the area is as large as possible.25. A man 2m tall walks at 3 m/s directly away from a streetlight that is 8 m high. How fast is the length of his
shadow changing?26. The hour and minute hands of a clock are 3 cm and 4 cm long respectively. How fast are their tips approaching
each other at 3 o'clock?
27. Obtain a cubic polynomial approximation to cos x at x = π3 and use it to approximate cos 57˚
28. A picture 2 m tall hangs on a vertical wall, the lower edge being 1 m above your eyes. How far from the wallshould you stand in order to obtain the "best view" of the picture? (i.e the angle subtended by the picture, at youreye, should be a maximum.)
29. Show that the curves xy = 2 and x2 - y2 = 1 intersect so that, at the point(s) of intersection, their tangentlines are perpendicular.
30. A tangent line, drawn to the curve x + y = 1, has x- and y-intercepts at P and Q. Show that the sum of theintercepts is a constant, independent of where the tangent line is drawn.
31. Determine where the polar curves r = q and r = cos q intersect, for q > 0. (You'll need to use Newton's method!)
32. Prove that π (1 - x) is always larger than 2x ln 1x (from the paint PARADOX) .
Hint: consider f(x) = π (1 - x) - 2x ln 1x . Is it always positive?
33. Evaluate the following integrals:
(a) ⌡⌠
0
∞
dx
1+x2 (b)
⌡⌠
1
∞
sin x
x dx (c) ⌡⌠
-∞
∞
e-|x| dx (d) ⌡⌠0
∞
arctan x dx
(e) ⌡⌠
-1
1
dx
x2 (f)
⌡⌠
0
πdx
x1/3 (g) ⌡
⌠
-1
1dxx-1 (h) ⌡⌠
0
1
ln x dx
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LECTURE 0
This is an introduction to the ideas of the calculus. Many students will have had a previous calculus course andbeen exposed to the idea of a limit, differentiation "rules" (product, quotient, chain rule, implicit differentiation,etc.), derivative of trig, exponential and log functions, optimization and related rate problems, some curvesketching, Riemann sums and the calculation of area. So, nearly all of this course should be vaguely familiar tomany students - but we won't count on it! In fact, we'll start from the beginning, assuming very little previousknowledge of the calculus ... so it'll be a review for many.
SOME BASICSNUMBERS ... and INFINITY:
Once upon a time there were only the positive integers: 1, 2, 3, 4, ... and we could add them or multiply themand we would again get positive integers: 17 + 21 = 38, 7 x 8 = 56.When we subtract however, we can get negative integers: 17 - 21 = -4, so we add these negative integers to ourcollection of numbers. Now we have all the integers (including 0) and when we add, subtract or multiplynumbers in this collection we get other numbers in the same collection.When we divide, however, we can get fractions which are not in our collection of positive and negative integers
... so we add them to our collection to get the set of "rational numbers". For example, 53 is a (rational) number
which, when multiplied by the integer 3 yields the integer 5.
What, then, is 50 ? If it's a number, then it should yield 5 when multiplied by 0. But every number yields 0 when
multiplied by 0 ... so 50 is NOT a number. Remember this! (It may be a cauliflower, but it's NOT a number!)
We will have occasion to "let x approach infinity", which we write: x->∞. This simply means that x is allowedto increase without bound; it becomes larger than ANY number.Is ∞ a number? No. It's a convenient symbol which we will use to indicate that a quantity (like x) is increasingso as to exceed (eventually) every number.
Consider the limiting value of the ratio 2xx+1 as x->∞. (We will have more to say about "limits" later.) The
numerator becomes infinite, as does the denominator. Can we write the limiting value as ∞∞ ... and is this equal
to 1? No. In fact, the limiting value of 2xx+1 is the number 2 (as x->∞). Similarly we can't say that ∞ - ∞ is 0
(as in the limiting value of x2 - x as x->∞). Since ∞ is NOT a number it's not surprising that it doesn't behavelike a number!
Remember:
0 x ∞ ≠ 0 and ∞ - ∞ ≠ 0 and ∞∞ ≠ 1 and
10 ≠ ∞ and
00 ≠ 1
PS:
S: Why isn't 00 = 1? I can accept
∞∞ ≠ 1 (since ∞ isn't a number then division isn't an operation one can perform on ∞) but 0
is a number.
P: Let' see. When multiplied by 0, 00 must yield 0. That's certainly true of the number 5 (that is, 5 multiplied by 0 does yield
0), so maybe 00 = 5. But then any number yields 0 when multiplied by 0. So maybe
00 can be any number. It certainly can't
be any specific number like 5. Let's say it's indeterminate.S: Fair enough. One last thing. Any number multiplied by 0 gives 0, right?P: Right.S: Then surely ∞ multiplied by 0 must give 0.P: And a brown cow multiplied by 0? Does it give 0? Remember, ∞ is NOT a number.S: Okay, okay. But are there any more numbers? I mean, if subtracting gave us the negative numbers and division the rational
numbers, why don't we do something else and get more numbers? Are there any more?
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P: Sure. Besides adding, subtracting, multiplying and dividing we can take roots, say the square root. That'd give us more
numbers. For example, 2 is not a rational number. It's irrational.S: And will taking roots give us all the numbers?P: No. There are numbers like π which can't be obtained by taking the root of a rational number.S: And what do you call these numbers?
P: They're irrational, just like 2 (because they aren't the ratio of two integers) , but they're also called transcendentalnumbers.
S: I've got a headache. Can we just keep going'?
INEQUALITIES:We will denote the set of numbers in the interval from -1 to 7, including both -1 and 7, as [-1,7] ... or sometimeswe'll use: -1 ≤ x ≤ 7. If -1 is included but 7 is not, we'll use the notation [-1,7) ... or sometimes -1 ≤ x < 7.If neither -1 nor 7 is included we'll use (-1,7) or perhaps -1 < x < 7.Intervals of the type a ≤ x ≤ b (which include both end-points) are called CLOSED intervals. Intervals of thetype a < x < b which include neither end-point are called OPEN intervals. The interval -1 < x ≤ 7 is neither opennor closed.
We are often required to "solve" inequalities like: 4x
3x+1 ≥ 1. Here, for example, we must find the values of x
which make 4x
3x+1 greater than, or equal to, 1. If we simply multiply both sides of the inequality by (3x+1), just
as we would with an equality (i.e. an "equation") we would get 4x ≥ 3x+1 hence x ≥ 1 (subtracting 3x from each
side) and we would conclude that only those values of x greater than (or equal to) 1 will make 4x
3x+1 ≥ 1. It is a
surprise, then, to find that x = - 10 also makes 4x
3x+1 ≥ 1. (Try it!)
The rule is this: when multiplying an inequality by a number (or an expression, like 3x+1) we must change thedirection of the inequality if the number (or expression) is negative.
Examples:• Although 7 > 5, after multiplying both sides by -3, we get -21 ≤ -15 (and the direction of the inequality has
changed). However, multiplying by 2 (a positive number) we get 14 ≥ 10 (and we don't change the direction ofthe inequality).
• If 4x
3x+1 ≥ 1, then 4x ≥ 3x+1 provided 3x+1 > 0. Hence, when we found that x ≥ 1, we were only finding x-
values satisfying 3x+1 > 0 (i.e. x > -1/3).To find all solutions we proceed as follows:
(i) We look for solutions satisfying 3x+1>0 (i.e. x > -1/3). Then 4x
3x+1 ≥ 1 is true provided 4x ≥ 3x+1, that is,
provided x ≥ 1. We now have all solutions satisfying 3x+1 > 0, namely all x-values satisfying x ≥ 1. But x > -1/3and x ≥ 1 means all x-values satisfying x ≥ 1.(ii) Now we look for solutions satisfying 3x+1<0 (i.e. x < -1/3). Multiplying the inequality by (3x+1) willchange its direction: 4x ≤ 3x+1, hence x ≤ 1. We now have, as solutions, all x-values satisfying x < -1/3 and
x ≤ 1 ... hence all x-values satisfying x < -1/3 (and that includes the solution x = -10 mentioned above!)
(ii) Finally, then, any value of x which satisfies either x ≥ 1 or x < -1/3 will satisfy 4x
3x+1 ≥ 1
Of course, a picture is worth a thousand words, so here's the
graph of y = 4x
3x+1 . Note that y is greater than or equal to 1 when
x <- 13 and then again when x ≥ 1.
FUNCTIONS:
If a quantity, which we'll call "y", depends upon another quantity (we'll call it "x") so that y assumes a single,unique value for each value of x (chosen from some set of values), then we say that "y is a function of x" and
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write y = f(x). The requirement of a "single, unique value" is important; if we don't have that, we don't have a"function". Graphically, this means a function must satisfy the vertical line test: every vertical line (in thedomain) must cross the graph of y = f(x) only once. If it crosses more than once, then there are two or more y-values and the graph is NOT that of a function.
Examples:• A person's height is a function of her age: height = f(age)• The volume of a sphere is a function of its radius: V = f(r)
The set of values from which x is chosen is called the domain of the function. The set of possible values for y iscalled the range of the function.
Note:It's common, when writing y = f(x), to call "x" the independent variable, "y" the dependent variable and "f" thefunction. The notation f(x) really means the value of the function "f" when the independent variable has thevalue x. In keeping with this notation we will sometimes refer to the function "f" (although we will sometimes,for the sake of clarity, refer to the function "f(x)"... unless confusing the value with the function makes thingsless clear!)
For example we might say "Consider the function x2" when we really mean "Consider the function which
squares" or perhaps "Consider the function which takes a number x and generates the number x2". The functionis really the operation (like "squaring"), not the result of applying this operation.Further, although we will often use "f" as a label for our function ... hence we'll refer to f(x) ... we will also useg(x) and h(x) etc. Curious how the labels for functions are often chosen from the middle of the alphabet: "f","g", "h" ... constants are chosen from the beginning: "a", "b", "c" ... and variables from the end: "x", "y", "z". Infact, if we're talking about temperature T as a function of time t, we might just write T(t), and if we're talkingabout pressure P as a function of volume V we might write P(V) ... and so on. In particular, it is commonpractice to write x(t) for the position of an object at time t and v(t) for its velocity.Let's talk a little about functions and their domain:
Examples:
• y is a function of x according to the rule: y = 1 - x2 , where the domain is the set of numbers -1 ≤ x ≤ 1 (else ywill be the square root of a negative number) and the range is the set: 0 ≤ y ≤ 1. (For each x in the domain, ywill lie in this range.). Sometimes the domain isn't specified explicitly, but is understood to be whatever x-values will provide a real value for y. In this example, we would identify the domain as -1 ≤ x ≤ 1 without beingtold!
• y = x2 +4x - 7. What is the domain? Since every real number x provides a real value for y, the domain is theentire real line: -∞ < x < ∞ (unless otherwise specified). The range is harder to identify. It's the totality ofpossible values of y, and happens to be -11 ≤ y < ∞. (Can you prove this?)
• Suppose the value of y is related to the x-value according to x2 + y2 = 25. In this example, if we write
y2 = 25 - x2 and recognize that y2, hence 25 - x2, must be positive (or, at least, non-negative) then we must have x2
≤ 25 so x must be restricted to the interval -5 ≤ x ≤ 5. However, for each x in this interval we can't guarantee that
the relation defines a single, unique y-value! For example, if x = 3, then y2 = 25 - 32 = 16 and the y-value could
be 4 or -4. Hence, the relation x2 + y2 = 25 does NOT define y as a "function" of x. We could, of course,
"solve for y" and obtain the possible "solutions": y = 1 - x2 or perhaps y = - 1 - x2 and either one will
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define a function with domain -1 ≤ x ≤ 1 ... but the original relation did NOT define a function.
ABSOLUTE VALUES: We use the notation | N | to mean the absolute value of the number (or expression) N. It's easy for numbers: | 7 |
= 7 and | -7 | = 7. In general, | N | = N if N is positive, and | N | = -N is N is negative, and | 0 | = 0.More formally, we define:
| N | = N if N ≥ 0- N if N < 0
and note that N can be a number (such as -7 or π ) or an expression (such as x2 - 3 or sin x).Examples:
• | x+1| = x+1 if x+1 ≥ 0-(x+1) if x+1 < 0 i.e |x+1| =
x+1 if x ≥ -1-(x+1) if x< -1
• | 4-x2| = 4-x2 if 4-x2 ≥ 0
-(4-x2) if 4-x2 < 0 i.e | 4-x2| =
4-x2 if x2 ≤ 4 (in -2≤x≤2)
-(4-x2) if x2> 4 (x<-2 or x>2)
• |x| + |y| = x + y if x ≥ 0 and y ≥ 0 (i.e. in the firstquadrant of the x-y plane, including the positive x- andy-axes), whereas |x| + |y| = x - y if x ≥ 0 but y < 0(the fourth quadrant) and |x| + |y| = -x - y in thethird quadrant and, finally, |x| + |y| = -x + y in thesecond quadrant.
The graph of |x| + |y| = 1 is shown at the right.
• To plot y = | f(x) |, just plot y = f(x) and reflect the negative parts in the x-axis (i.e. replace negative values off(x) by - f(x))
Final Note: Now, with both absolute values and inequalities covered, we can consider expressions like
| x2 - 4 | < 2, which is the same as - 2 < x2 - 4 < 2 or 2 < x2 < 6 hence x lies in the interval 2 < x < 6 OR in
- 6 < x < - 2 . This manipulation is important when we come to consider expressions like| x - a | < h (which is the same as a - h < x < a + h) or the expression | f(x) - L | < e (which is the same as
L - e < f(x) < L + e).• Reminder: | N | < 3 means -3 < N < 3 and
| x+5 | < 3 means -3 < x+5 < 3 (hence -8 < x < -2)etc. etc. etc.
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SOME TRIG IDENTITIES and other trig stuff* :
• sin2A + cos2A = 1 for any number (or angle) A.• sin (A+B) = sin A cos B + cos A sin B and sin (A-B) = sin A cos B - cos A sin B• cos (A+B) = cos A cos B - sin A sin B and cos (A-B) = cos A cos B + sin A sin B
• sin A - sin B = 2 cos A+B2 sin
A-B2
• sin2 A = 1 - cos 2A
2 and cos2 A = 1 + cos 2A
2
Unless otherwise specified, if we write sin A , cos A, etc., and if you wish toconsider A an angle, then assume it's the RADIAN measure of an angle!
• sin A and cos A are defined for real numbers "A" as follows (and you'll note that the definition has little to dowith angles!)
Given any positive number A, begin at the point (1,0) on the unitcircle and move counter-clockwise a distance equal to A,reaching a point we can call P. The x-coordinate of P is cos Aand the y-coordinate of P is sin A (and that's the definition of thesine and cosine of a number A). If A is negative, then moveclockwise.
We can now introduce an angle "A". Note that the angle at the centre of the circle has the radian measure A ... since
arclength=radius x angle and since our radius is "1", the arclength and the central angle are equal ... provided
the central angle is measured in RADIANS! We're all familiar with circumference = 2πr which is the above
formula with the central angle a complete revolution: 2π, in RADIANS! We'd never write circumference = 360r, where the central angle is 360 when measured in degrees, so we restrict ourselves to RADIANS when weconsider angles (unless, of course, we indicate otherwise).
Anyway, because of the definition we can easily identify a number ofpoints on the unit circle ... like (1,0) and (0,1) and (-1,0) and (0,-1) ... hence the value of sin A and cos A for various numbers A:
sin 0 = 0 and cos 0 = 1
sin π/2 = 1 and cos π/2 = 0
sin π = 0 and cos π = -1
sin 3π/2 = -1 and cos 3π/2 = 0
PS:
S: Can't I write cos 45 = 1
2 ?
P: Sure. Just be sure you indicate that the angle is in degrees, like cos 45˚ = 1
2 , else somebody will measure off 45 units
along the circle and take the x-coordinate of the resultant point as cos 45 ... and it won't be 1
2 .
S: You gotta be kiddin'. Everybody knows cos 45 and sin 90 etc. are in degrees.P: And if I write cos 47?S: No, I mean the standard angles like 30, 45, 60, 90 and so on ... not 47.P: Wait till we get into course. We'll approximate cos 47, knowing cos 45, and we'll all be confused if we don't agree on this.
* Leonard Euler (1707-1783) is responsible for the modern treatment of logarithms and exponential functions andintroduced the notations sin(x), cos(x), etc. and f(x). A Swiss mathematician, he studied math under JohannBernoulli but soon outstripped his teacher.
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S: Okay ... radians it is. But how about the good ol' sin A = opposite
hypotenuse . Has that gone?
P: No, that still works. We take some angle A and stare at the coordinates of the point P.
See the triangle? Notice that opposite
hypotenuse = sin A. (Of course, the hypotenuse is "1" and the "opposite" is just the y-
coordinate which, by definition, is sin A).S: Seems like cheating. Your hypotenuse is always "1". Is that necessary?P: Let's enlarge the whole diagram ... blow it up ... like a photographic enlargement,
by a factor R. Then we have the following diagram. See? The triangle has
hypotenuse R and opposite
hypotenuse =
R sin AR
= sin A, again, and of course,
adjacenthypotenuse
= R cos A
R = cos A. Happy?
S: No.
SOME TRIG GRAPHS to remember:
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S: How about the special angles, 45˚ and 30˚ and so on.P: There's nothing special about them, except that most people remember
the 6 trig functions for 30˚, 45˚ and 60˚ from the triangles shown at theright ===>>>
Then sin 45˚ (or sin π4 ) =
1
2 and cos 60˚ (or cos
π3 ) =
12 and cos 30˚
(or cos π6 ) =
32 and, of course, knowing these triangles you can read
off the other trig functions as well, like tan 60˚ (or tan π3 ) = 3 and
so on. These will come in handy when we have to sketch the graph of curves involving angles.
SOME GEOMETRY: All straight lines have an equation of the form: Ax + By = C.
The equation of a line through the points (x1,y1) and (x2,y2) is: y - y1x - x1
= y2 - y1x2 - x1
.
The equation of a line through (a,b) with slope m is: y - bx - a = m (the point-slope form).
If m1 and m2 are the slopes of two lines, then they will be perpendicular if m1 m2 = -1.
LOGARITHMS and EXPONENTIALS:
The functions 2x, 5x, πx or ax (for any positive number "a") are called exponential functions, and "a" is calledthe base. There are RULES:
(A) ax ay = ax+yax
ay = ax-y ( )ax
n = anx a0 = 1
If ay = x, then y is called the logarithm of x to the base a, written: y = loga x, a logarithmic function (or simply
a log function). There are RULES:
(B) log xy = log x+log y log xy = log x-log y log xn = n log x log 1 = 0
Every entry in (B) actually follows from the corresponding entry in (A). For example, assume the logs in (B) are
to the base "a". Then alog x + log y = alog x alog y = xy = alog xy hence we get the first entry in (B), where
we've used the fact that alog N = N if the log-base is "a".
Notice that x = ay is an exponential function of y, and if we solve for y we get y = loga x, a log function of x. In
fact, x = ay and y = loga x are two different ways of writing the same relation between x and y. (i.e. if x = ay
then y = loga x AND if y = loga x then x = ay.)
Hence we may write x = alogax or even y = loga a
y .
Examples:
• log3 37 = 7 and log5 5 = 1 and logπ1 = 0
• 3log37 = 7 and 56 = 2
6log25 and π0 = 1PS:S: That's confusing ... isn't it? I was never very good at logs.P: Everybody loves exponentials. Nobody loves logs. Everybody knows the RULES (A). Nobody knows (B). Well ... some
don't, and the examples are even more unfriendly, right?S: Right! So should I memorize (B)?P: Yes ... and remember: a "log" is an "exponent" and all properties of logs follow from the rules for exponents ... the log-rules
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just look different, but they're really familiar things.S: Sure, sure.
MORE ON LOGS:
If p = logaN, then ap = N. Now we take logs of each side, to the base b: logb(a
p) = logbN. Using a magic
property of logs (to any base) we get: p logb a = logbN, hence p = logbN
logb a . Remembering who "p" is, we
have: logaN = logbN
logb a . So, we can easily change logs from one base to another. If you have a table of logs to the
base 10 and desperately need logs to the base 2, just use log2N =
Error!. The above relation has a cousin: let N = b and recall that logb b = 1. Then we get: Error!which will beuseful when we consider log functions later on.
PS:S: Wait just one minute. Do you really expect me to remember all this .. all this ...P: No. But it's here so you can look it up when you need it ... and someday you may need it.S: So I only have to know that there are umpteen weird relations and I should know where to look to find them?P: Yes.S: You said log properties are familiar things, but I wouldn't call logab = 1/logba a familiar thing.
P: Pay attention: if x = logab then ax = b, and if y = logba then b
y = a and we want to show that x = 1/y, right? Okay, if ax = b
then raise both sides to the power y and get (ax)y = by , but by = a, so axy = a and that means that xy = 1. See? It's just aproperty of exponents. Right?
S: zzzzzz
ODDS 'n' ENDS ... mostly ENDS:
• A geometric series has the form: a + a r + a r2 + a r3 + ... + a rn-1 (there are n terms here ... count 'em!).
The sum of the above series is a 1 - rn
1 - r . If | r | < 1, then
limn->∞
a 1 - rn
1 - r = a
1 - r since rn->0.
That is, the sum of the "infinite" geometric series is a + a r + a r2 + a r3 + ... = a
1 - r provided | r | < 1!!
• Polynomials are functions like 1 + x - 3x2 or x3 - 5x4. They all have the form A + Bx + Cx2 + Dx3 + ... and soon, ending after a finite number of such terms. "A", "B", etc. are constants. Further, if p(x) and q(x) are
polynomials, then p(x)q(x) is called a rational function.
PS:S: Whoops. Has that anything to do with rational numbers?
P: It's the same kind of definition. If p and q are integers, then pq is a rational number. If p and q are polynomials, then
pq is a
rational function. See? The polynomial functions play the role of the integers.S: Sure, sure.
• SIGMA NOTATION:We will have occasion to refer to a series, say a1 + a2 + a3 + ... + an, and it gets tiring to have to write out
several terms (as we have just done) every time we want to identify the series. For this reason we may useSIGMA notation: each term of the series has the form ak where k = 1 or 2 or 3 ... or n, and the series is the sum
of such terms, so we write ∑k=1
n ak where ∑
k=1
n means "sum all such terms from k = 1 to k = n". For example:
1 + 2 + 3 + ... + 100 = ∑k=1
100 k and ∑
k=7
12
ek = e7 + e8 + e9 + e10 + e11 + e12 and
12
f(1+h) + f(1+2h) + f(1+3h) + ... + f(1+nh) = ∑k=1
n f(1+kh) and ∑
p=4
5sin πp = sin 4π + sin 5π = 0
and ∑i=0
∞
ri = 1
1 - r (provided | r | < 1) .
PS: The Greek letter ∑ is called SIGMA ... it's upper case ... lower case sigma is s.
• In these lectures we will, from time to time, need to do some algebra and arithmetic with lots of digits of
accuracy (just to illustrate some point we're making; we won't expect you to do these calculations). Then we'll
use a computer program called MAPLE (except we'll call it 'cause it looks nicer). is a
so-called Computer Algebra System (or CAS) which knows many of the techniques we'll learn in this course
and, in particular, can do arithmetic with great (infinite?) precision.
LECTURE 1
LIMITSPS.P: What's the limit of the sequence of numbers: 4.9, 4.99, 4.999, 4.9999, etc. etc. ?S: It's 5.P: Why 5?S: Because the numbers get closer and closer to 5.P: They also get closer and closer to 17 ... so why isn't the limit 17?S: Well ... they get closer to 5 than to 17.P: But one of the numbers is exactly 4.99, so maybe that's the limit. After all, how much closer can you get?S: Oh, they have to get closer and closer without actually reaching the limit ... so that's why it's 5 and not 4.99 ... I think.P: How about the following sequence: 4.9, 5, 4.99, 5, 4.999, 5, 4.9999, 5, etc. etc. where every second number is exactly 5.
Now what's the limit? It's still 5, right? Yet the numbers actually reach 5 from time to time. Clearly we need a definition ofwhat we mean by "the limit is 5" so that only one number satisfies the definition and that one number is 5 ... and not 4.99 or17.
S: If you say so ... but what good is all this?P: We want to be able to describe, in a reasonably precise manner, the various features of a graph such as:
S: You're kidding. Surely these functions never occur in a real problem, right?
13
P: Well, we want to establish a kind ofvocabulary so we can talk about thefeatures ... with a terminology that hassome precise meaning. Of course, if wecan analyze the above weird function thensomething like ==>>is really easy.
S: What does this function represent?P: It could be the concentration of a drug in
the bloodstream as a function of time, orthe amount of energy radiated by a hotbody as a function of the wavelength ofthe emitted light, or some probabilitydistribution or maybe ...
S: Okay, okay, but what would one want to know about such a function?P: Where the maximum occurs or what the limiting value of y is when x becomes infinite or how y behaves for small values of
x or ..S: Okay, let's go.
Consider the function f(x) = 4x2-92x-3 . Notice that f(x) has no value at x =1.5 since f(1.5) gives
00 which is not a
number. Nevertheless, f(x) does have a limiting value of 6. (We make this claim without even defining what wemean by limiting value! We'll provide this definition below in such a way that the mathematical definition agreeswith our common sense notion of limiting value. First, then, we need to think about this common sense notion.)To see that f(x) does, indeed, have a limit of 6, we compute f(x) for various values of x approaching 1.5, and wealso include the error, namely |f(x) - 6|, the absolute value of f(x) - 6 (to see how well we're doing in achievingthe limit of 6).
Table 1 Table 2
x f(x) error
1.3 5.6 0.4
1.4 5.8 0.2
1.49 5.98 0.02
1.499 5.998 0.002
1.4999 5.9998 0.0002
x f(x) error
1.7 6.4 0.4
1.6 6.2 0.2
1.51 6.02 0.02
1.501 6.002 0.002
1.5001 6.0002 0.0002
In spite of the fact that f(1.5) doesn't exist, it seems clear from the table that the limit (as x approaches 3/2) is 6. (We
write this as:
lim
x->3/2 f(x) = 6.) Indeed, by making x sufficiently close to 1.5, we can make the error, namely
|f(x) - 6|, as small as we please. For example, suppose we want the error to be less than, say, .001 (meaning wewant the values of f(x) to lie in 5.999 < f(x) < 6.001). We can achieve this simply by making x sufficiently closeto the number 1.5 (for example, we can restrict x to lie in the interval 1.4999 < x < 1.5001).For this particular function and this particular x-value, we can say:Since the error, |f(x) - 6|, can be made as small as we please simply by restricting x to lie in some sufficientlysmall interval about the number 1.5, then the limit of f(x) is 6, as x approaches 1.5 and this is indicated by the
notation:
lim
x->3/2 f(x) = 6.
S: That's confusing. I mean, it seems easy enough to do ... this business of making something small.P: Not just small but as small as we please, and remember that we should be able to do this without prescribing the value of
f(x) or even the value of x but only by controlling how far x is from the number 1.5, and it's precisely this property we wantfor our "limit" definition, and ...
S: So why does that make 6 the limit and not, say, 7?
We can't make the error, |f(x) - 7|, as small as we please by restricting x to lie in some small interval about 1.5,because when x is close to 1.5 the values of f(x) are close to 6, NOT close to 7 ... so 7 is NOT the limit. In fact,6 is the ONLY number that satisfies this definition!
14
S: I think you have a problem there. I can make x really close to 1.5 by making it EQUAL to 1.5, then is your "error" small? I
mean, how small is |f(1.5) - 6|? I make it | 00 - 6 | which isn't even a number, let alone a small number .
P: Very good! You've put your finger on a problem with our definition ... so lets modify it.
Alas, no matter what interval we choose about 1.5, x = 1.5 will be in that interval ... and there's no way we canmake the error small for x = 1.5 since f(x) doesn't even have a value! To fix this, we modify our definition toread:
We say that the limit of f(x) is 6, as x approaches 32 (i.e.
lim
x->3/2 f(x) = 6) if the error, |f(x) - 6|, can be made as
small as we please simply by restricting x to lie in some sufficiently small interval about the number 1.5, withthe exception of x =1.5 itself!
S: Don't you find that awkward? I mean, "by restricting x to lie in some sufficiently small interval about the number 1.5, withthe exception of x =1.5 itself" sound like mumbo-jumbo.
P: Okay, let's modify it ... again. We have to find something to replace that phrase which, I admit, is rather awkward, but I'mnot sure you'll like the modification.
We can improve upon the wording (at the expense, perhaps, of making it less understandable!) by replacing thephrase "by restricting x to lie in some sufficiently small interval about the number 1.5, with the exception of x=1.5 itself" with the phrase "by choosing a sufficiently small number, h, and restricting x to lie in the interval, 0< |x - 1.5| < h". Note that |x - 1.5| < h means that 1.5 - h < x < 1.5 + h, so we indeed have restricted x to someinterval about 1.5 ... and we can make the interval small by choosing a small h ... and 0 < |x - 1.5| means that
x ≠ 1.5 ... so we've now got a reasonable definition of limit.We'll generalize this notion of limit to other functions, and other x-values:
We say that
limx->a
f(x) = L if we can make the error, |f(x) - L|, as small as we please, simply by
choosing a sufficiently small number, h, and restricting x to lie in the interval, 0 < |x - a| < h.
S: You're right. I don't like the mods you've made. Can we go back to the original?P: No. We'll leave it as it is. Don't you see how pretty it is? You just write 0 < |x - 1.5 | < .0001 and you've said that x is not
equal to 1.5, but it's very close. Now pay attention.
Now let's return to the function f(x) which we can write as 4x2-92x-3 =
(2x-3)(2x+3)2x-3 = 2x+3 provided 2x-3≠0. But
if we're considering the limit as x->3/2, then x is close to but different from 3/2, so the division of numerator anddenominator by 2x-3 is valid ... and it's clear why f(x)->6 as x->3/2 (since f(x) is identical to 2x+3 for every x-value except x = 3/2 ... so the limit of f(x) is the same as the limit of 2x+3, namely 6).
Often, we can avoid using the above definition to find a limit (assuming a limit exists). In fact, from thedefinition, one can prove some nice rules (but we'll omit the proofs).
LIMIT RULES
Suppose
limx->a
f(x) = L and
limx->a
g(x) = M. Then we have the following rules:
SUM RULE:
limx->a
( )f(x) + g(x) = L + M DIFFERENCE RULE:
limx->a
( )f(x) - g(x) = L - M
PRODUCT RULE:
limx->a
f(x) g(x) = L M QUOTIENT RULE:
limx->a
f(x)g(x) =
LM (provided M ≠ 0)
15
Example: If f(x) = x2
x-3 , calculate
limx->2
f(x).
Solution: As x->2 we have lim f(x) = lim x2
x-3 = lim(x2)lim(x-3) =
lim(x) lim(x)lim(x) - lim(3) =
(2)(2)(2)-3 = - 4.
Note that the value of this function, namely f(2) (which we get by direct substitution), is also -4.It's important to notice, however, that we did NOT evaluate the limit by substituting x = 2. Instead we used thevarious LIMIT RULES. If the limit turns out to be f(2) it's because we have a CONTINUOUS function (seebelow for a discussion of "continuous" functions).
PS:S: Hold on! It looks to me like you just plugged in x = 2 to get the limit, right? And that gives you the value, right? And that
makes the limit equal to the value, right?P: Wrong. I used the various LIMIT RULES. It's just happens that the result is -4, the same as the value. But that's only true for
certain functions ... called "continuous" functions. It's not always true that the limit and the value are equal.S: Example?P: Okay, let's see ... I can invent a function which has no value but does have a limit. Want to see it?S: Sure.
P: It's f(x) = 4x2-92x-3
which has no value at x = 32 but has a limit of 6. Like it?
S: Yeah, it's great. But how about a function which actually does have a value. Then it's the same as the limit, right?P: Wrong. I'll invent a function defined for every real number x, and it will then have a value at x = 1.5, but this function will
also have a limit as x->1.5 and it'll be different. Want to see it?S: Sure.
P: It's f(x) = 4x2-92x-3
for all x ≠ 1.5 and f(1.5) = 47, or I could write it as: f(x) =
4x2-92x-3
if x ≠ 1.5
47 if x = 1.5
. Like it?
S: Hey! I don't mean a double-barrelled function! I mean ...
P: Okay, here's another one: f(x) = x2 + x2
(1+x2) +
x2
(1+x2)2 +
x2
(1+x2)3 +
x2
(1+x2)4 + ... Like it?
S: No. I assume it goes on forever. I don't know anything about functions that go on forever. I mean ...P: Not true. You know how to find the sum of such an infinite series as this. Look at it. It's a geometric series and the common
ratio is ...
S: Wait, I'll do it. The common ratio is ... uh, I divide the second term by the first and I get ... uh, 1
1+x2 . Right?
P: Sure, but you should also divide the third by the second and the fourth by the ...S: I know that, but you said it was geometric. Anyway, I can only add an infinite geometric series if the common ratio is less
than 1 and ... uh, well, I guess 1
1+x2 is less than 1. Terrific. Then it adds up to
a1 - r
and that's x2
1 - 1
1+x2
and that's ... uh,
1 + x2. So what?
P: So find
limx->0
f(x).
S: I guess it's
limx->0
(1 + x2) = 1. So that's the same as f(0), right? I mean, f(x) = 1 + x2 so f(0) = 1.
P: Wrong! When x ≠ 0, then your common ratio is less than 1 so you can add the infinite series using a
1 - r , BUT when x = 0
your common ratio is exactly 1 so you CAN'T use this formula.S: Then how do I get f(0)? Wait! I just plug it in! f(0) = 0 + 0 + 0 + ... which I guess is 0, right?P: Right! So for this function the limit is "1" but the value is "0". Nice, eh?S: No. Anyway, I read, somewhere, the following definition of "limit":
16
limx->a
f(x) means:
for any e > 0, a d can be found such that | f(x) - L | < e whenever 0 < | x - a | < d
P: Yes, it's the same definition as ours. When we say we can "make the error as small as we please" we really mean "smallerthan any number e > 0", and when we choose an "h" and "restrict x to lie in the 0 < |x - a | < h", it's the same as finding a "d"and restricting x to lie in the interval 0 < | x - a | < d. See? It's the same. Don't let the Greek letters fool you.
S: Sure, sure. So why even use this definition?
P: Suppose
limx->3
f(x) = 47. Then, by restricting x to a sufficiently small interval about "3", say 0 < | x - 3 | < d, we can make
the error | f(x) - 47 | < .1, say. Hence we can force f(x) to lie in the interval 46.9 < f(x) < 47.1, and if we want the error to beeven smaller, say | f(x) - 47 | < .001, then we can select a smaller value for d, and ...
S: Wait. You've already said all this. I asked why the e - d definition is any better than our earlier definition.P: I guess it's because we've given a name to the error, namely e which could be .1 or .001, etc.S: I still don't see what good all this is. Is there any useful application of this stuff ... something I'd understand?
P: Well ... suppose you were building a box with all sides of equal length and it was to have a volume of 8 m3 with an error of,
say, 0.1 m3 (which is e by the way). Then each side would be about 2 metres (since 23 = 8). Now, the big question: how
accurately must you cut the sides so that the error in the volume is less than 0.1 m3? We write V(x) = x3 and insist that theerror in volume, | V(x) - 8 |, is less than 0.1 by restricting x to lie in some interval about x = 2, say 2 - h < x < 2 + h. Theanswer to the question above is the value of "h" ... so how small must h be?
S: Let me do it! I'd want V(x) to lie between 8 - 0.1 = 7.9 and 8 + 0.1 = 8.1 ('cause that'd make the error less than 0.1 m3). That
means I'd want 7.9 < x3 < 8.1, so I'd take the cube root and find that 1.9992 < x < 2.00083, so I'd have to cut the sides withan error less than ... uh, about .0008 m, right?
P: Right. You can guarantee an volume error less than 0.1 m3 by making the side length lie in 2 - h < x < 2 + h, withh = .0008 (i.e. x must lie in 1.9992 < x < 2.0008). Of course, if somebody wanted the volume error even smaller, less than
e = 0.001 m3 for example, then you'd have to choose an even smaller value of h.
S: That doesn't sound like calculus to me ... just common sense and a little arithmetic.P: Here's the calculus: the reason you can guarantee that an "h" exists for every specified error (which we can call "e") is
because the function V(x) = x3 is a continuous function of x, hence has a value and a limit at x = 2 and they're the same. Ifthis weren't the case then you couldn't guarantee an arbitrarily small error.
S: For example? I mean, are there really problems like that ? I mean real-world problems, not mathematical problems.P: Sure. Suppose the cost of postage is determined by the weight of the parcel according to the prescription:
cost = $10.00 for parcels under 1 kg, $15.00 for parcels from 1 kg to less than 2 kg, $20.00 for parcels from 2 kg to less than3 kg, etc. etc.If x is the weight and C(x) the cost, the graph looks like this ==>
Note that
lim
x->1.5 C(x) =15 , and because this limit exists we can guarantee the cost
to be near $15 (in fact, in this problem, exactly $15) by restricting the weight tolie in some sufficiently small interval about 1.5 kg. However,
limx->2
C(x) doesn't exist so it's NOT possible to guarantee a cost near $15 or near $20 (or near any amount L) by restricting
the weight to lie in some interval about 2 kg. In fact, a wee bit under 2 kg and the cost is $15 and a wee bit over 2 kg and thecost jumps to ...
S: Okay ... I got it ... calculus is wonderful.P: One more thing. Most students are wary of double-barrelled functions defined like:
f(x) = x2 if x < 0 x+1 if x ≥ 0
. In fact, there is a feeling that only mathematicians could love such functions. But the cost of
postage is given by C(x) = 10 if 0<x<1 15 if 1≤x<2 20 if 2≤x<3
and we could add to this array for parcels where x≥3 kg. Can't you just see
such a sign hanging in the post-office?S: No.
17
ONE SIDED LIMITS:
The graph of y = f(x) = 4x2-92x-3 is shown. Note that f(x) has no
value at x = 1.5 (which we indicate with an "open circle"),however, as x approaches 1.5 either from the left (i.e throughvalues such as 1.49, 1.499, 1.4999, etc. as in Table 1 above) orfrom the right (through values such as 1.51, 1.501, 1.5001, etc.as in Table 2), the values of f(x) clearly have the limit 6.
Now consider the graph of another function which we'll call g(x).This function does have a value at x = 1.5, namely g(1.5) = 6(indicated by the "closed circle"), however, if x approaches 1.5from the right, the values of g(x) have a limit of 7. We indicate
this by writing
lim
x->1.5+ g(x) = 7 where the notation x->1.5+
means x is approaching 1.5 through values more positive than1.5.
Further, if x approaches 1.5 from the left the limiting value of g(x) is 6, and we can write
lim
x->1.5- g(x) = 6 where
x->1.5- means x is approaching 1.5 through values more negative than 1.5.The question we pose is: does g(x) have a limit as x approaches 1.5?The answer will be "yes" only if our definition of limit is satisfied. If, for example, we suspect that
lim
x->1.5 g(x) = 6, then we must be able to make the error, |g(x) - 6|, as small as we please by restricting x to lie in
some interval about 1.5 such as: 0 < |x - 1.5| < h. This is clearly impossible since an x-value just slightly larger
than 1.5 will give g(x) a value near 7 so the error is already larger than 1.0, hence
lim
x->1.5 g(x) = 6 is NOT true.
Similarly,
lim
x->1.5 g(x) = 7 isn't true either (since values of x close to 1.5 but slightly smaller will give an error larger
than 1.0). Indeed, there is NO number L such that |g(x) - L| can be made as small as we please by restricting x to
lie in an interval about 1.5, hence
lim
x->1.5 g(x) = L is NOT true and we conclude that
lim
x->1.5 g(x) doesn't exist
(meaning no number L will satisfy our definition of limit).
In this example, g(x) has a right-sided limit,
lim
x->1.5+ g(x) = 7, and a left-sided limit,
lim
x->1.5- g(x) = 6, but it
doesn't have a limit:
lim
x->1.5 g(x). On the other hand, for the function f(x) =
4x2-92x-3 we have:
18
lim
x->1.5+ f(x) =
lim
x->1.5- f(x) =
lim
x->1.5 f(x) = 6.
That is, it has right- and left-sided limits and they're identical and equal to the limit. We make note of this:
limx->a
f(x) = L provided
lim
x->a- f(x) = L and
lim
x->a+ f(x) = L .
Also,
if
lim
x->a- f(x) = L and
lim
x->a+ f(x) = L then
limx->a
f(x) = L.
In words:
If
limx->a
f(x) = L exists, then both left- and right-sided limits will be equal to L.
Also, if both left- and right-sided limits exist and are equal to L, then L is also the limit of f(x).
Sometimes it's necessary to evaluate both left- and the right- limits in order to see if the limit exists.
Example: Does f(x) = |x2 - 9|x - 3 have a limit as x approaches 3?
(Or, to put it differently, does
limx->3
|x2 - 9|x - 3 exist?)
Note: When we see an absolute value sign we desperately want to get rid of it by using the fact that
|m| = m if m ≥ 0 whereas
|m| = - m if m < 0
(where m is any expression, such as x2-9).
Solution: To rid ourselves of the absolute value sign we need to know whether x2 - 9 is positive or negative,
so we first consider the limit as x approaches 3 from the left (i.e. x->3-). Then, for x slightly smaller than 9 (such
as 8.9 or 8.99 etc.), x2-9 < 0 so |x2 - 9| = - (x2-9). (Recall the definition of absolute value!)
Hence:
lim
x->3-
|x2 - 9|
x - 3 =
lim
x->3-
-(x2 - 9)
x - 3
=
lim
x->3-
-(x - 3)(x + 3)
x - 3 =
lim
x->3-( )- (x + 3) = - 6.
On the other hand, if we approach from the right
(so that x2 - 9 > 0 and |x2 - 9| = x2 - 9) we have:
lim
x->3+
|x2 - 9|
x - 3 =
lim
x->3+
(x2 - 9)
x - 3 =
lim
x->3+
(x - 3)(x + 3)
x - 3 =
lim
x->3+ (x + 3) = 6.
Since these limits are different, we conclude that
limx->3
|x2 - 9|x - 3 does NOT exist.
In fact, to the left of x = 3, the graph of y = |x2 - 9|x - 3 is the same as the graph of y = - (x + 3) and, to the right of
x = 3, the same as y = x + 3 as shown in the diagram. Clearly, left- and right-handed limits are NOT equal.
19
PS.S: If g(x) has a limit as x approaches 3 from the left, and it also has a limit as x approaches 3 from the right, then it has a limit
no matter how x approaches 3 (they ain't no other way to approach 3, is there?) So how come you say there's no limit as xapproaches 3?
P: It's that neat definition that does it. In order to have a limit (according to our definition) there must be some number L suchthat the error |g(x)-L| can be made as small as we please simply by forcing x to lie in some small interval about 3, such as 3-h<x<3+h (except for x=3 itself, of course). But it's impossible to find such a number L and such an interval because if x is inthe right part of the interval, 3 < x < 3+h, then g(x) can be made as close as we please to 6 ... so clearly L (whoever he is)must be VERY close to 6. On the other hand if x is in the left half of this interval, 3-h<x<3, the values of g(x) can be madeas close as we please to -6 hence L must be VERY close to -6 as well. But there's no number L which is simultaneouslyVERY close to 6 and VERY close to -6 ... so there's no L, hence there's no limit. See?
S: Not really.P: Don't worry about it ... it's not something we'll spend much time on. One of these days you'll wake up and run naked thru'
the streets shouting eureka! ... and all will be clear.S: (He's kidding ... right?)
LECTURE 2
INFINITE LIMITS, ASYMPTOTES and CONTINUOUS FUNCTIONS
INFINITE LIMITS:
We want to be able to describe the features of the above graph at x = -1 and x = 5. To do this we first consider
the limit:
limx->0
1
x2 . For x in some small interval about x = 0 (say -.001 < x < .001, excluding x = 0 itself) the
values of 1
x2 are larger than
1
(.001)2 = 1,000,000. Further, as we make the interval even smaller, the values of
1
x2 become even larger. In fact, we can make the values of this function larger than any number you care to
mention - and certainly larger than L for any number L - hence the the limit cannot be L for any L - hence there
is no limit. i.e.
limx->0
1
x2 does not exist.
Now consider
limx->0
sin 1x . This limit won't exist either, but for quite a different reason.
For x in some small interval about x = 0 (say -.001 < x < .001, excluding x = 0 itself) the values of sin 1x will be
between -1 and +1 (that's because the sine function always lies between -1 and +1). Furthermore, no matter how
20
small we make the interval we can always find x-values in the interval such that 1x is a multiple of π and that
will make sin 1x = 0 (since the sine of a multiple of π is 0) ... so, if
limx->0
sin 1x did exist and is equal to some
number L, then L must clearly be VERY close to 0. But, no matter how small the interval we can also find x-
values such that 1x is an odd multiple of
π2 and that will make sin
1x = ±1 (since the sine of an odd multiple of
π2 is either +1 or -1). Conclusion? sin
1x takes on values -1 and 0 and +1 in our interval (no matter what
interval we choose!) so the limit L (if there is such a number) must be close to -1 and close to 0 and close to +1,
all at the same time - clearly there is no such number - hence there is no limit. i.e.
limx->0
sin 1x does not exist.
PS.
S: Wait a minute. You say: we can always find an x-value such that 1x is a multiple of π ... what does that mean?
P: Pick an interval about x = 0.S: How about -.01 < x < .01?
P: Good. Now watch me find an x-value in your interval which will make sin 1x = 0. That means I have to find an x such that
1x
is a multiple of π ('cause that'll make sin 1x = 0). Let's see, if I choose
1x = 1000π (that's a multiple of π, right?) that'll mean
x = 1
1000π and since π is roughly 3 then x is roughly 1/3000 or roughly .0003 so my x certainly lies in your interval. And
believe me, if you choose any other interval about x = 0 I'll still be able to find x-values which will make sin 1x = 0. Not
only that, for any interval you choose I can also find an x-value such that sin 1x = 1. Not only that ...
S: Wait, wait ... let's see you find an x in -.0001 < x < .0001 which makes sin 1x = 1.
P: Okay, I'll need 1x to be
π
2 + a multiple of 2π ('cause that'll make sin
1x = 1) and I'll choose a huge multiple of 2π (so my x
lies in your interval) so let's try 1x =
π2 + 100,000(2π) (which makes sin
1x = 1) and we'll see if it's in your interval ... π is
roughly 3 so 1x is roughly 1.5 + 600,000 which is roughly 600000 so x is roughly
1600000
which is roughly .0000017 so
my x does indeed lie in your interval, and if you choose a smaller interval I'll just choose a bigger multiple of 2π and my xwould still lie ...
S: Let's forget the whole thing.
P: Wait. Don't you see? You can't possibly make the values of sin 1x as close as you please to some number L by restricting x
to lie in some interval about x = 0. Why? Because I can find x-values which make sin 1x equal to 1 and other x-values
which make sin 1x equal to 0 (to pick two convenient values of the sine function, though I could pick others ...), so L
(whoever she is) must be VERY close to 1 and VERY close to 0 all at the same time ... and there is no such number. In fact,
since sin 1x takes on every value between -1 and +1 in every interval about x=0, the limit L must be close to every number
between -1 and +1, all at the same time ... and that's impossible ... so there is no limit L. Understand?S: zzzzz
Neither of the limits
limx->0
1
x2 and
limx->0
sin 1x exist (because there is no number, L, which satisfies our
definition of a "limit"). Yet, the first fails to exist for a particular reason, namely: the values of 1
x2 become
larger than any number when x is restricted to smaller and smaller intervals about x = 0 (excluding, of course,
21
x = 0 itself). In order to indicate this particular reason for nonexistence of a limit we write:
limx->0
1
x2 = ∞.
Writing this doesn't mean that we now have a limiting value for 1
x2 . It simply means that the limit fails to exist
because 1
x2 becomes larger than any number as x approaches 0, or, to put it differently, we say:
1
x2 approaches
infinity as x approaches zero. In general:
limx->a
f(x) = ∞
means that the values of f(x) can be made arbitrarily large by restricting x to lie in some sufficiently small interval about x = a, excluding x = a itself.
or, in sexier words,
limx->a
f(x) = ∞
means that the values of f(x) can be made arbitrarily large by choosing a sufficiently small number h and restricting x to lie in the interval 0 < |x - a | < h.
It should be clear what we meaning by:
limx->0
-1
x2 = -∞
We also have (think about these!):
lim
x->0+ 1x = ∞ and
lim
x->0- 1x = -∞ and
lim
x->1+ 1
x2-1 = ∞ and
lim
x->1- 1
x2-1 = -∞
Note that we use ∞ to mean +∞ (the + is understood, just like 5 means +5).
ASYMPTOTES:
If any of the following are true, then we say that the graph of y = f(x) has aVERTICAL ASYMPTOTE, namely: x = a.
lim
x->a- f(x) = ∞
lim
x->a- f(x) = -∞
lim
x->a+ f(x) = -∞
lim
x->a+ f(x) = ∞
22
Examples: Evaluate the following limits (or explain why they don't exist)
(a)
limx->0
1x (b)
lim
x->3+
x
9 - x2
(c)
lim
x->0+ 2
1/x(d)
lim
x->0- 2
1/x
Solutions: (a)
lim
x->0- 1x = -∞ whereas
lim
x->0+ 1x = ∞, and they're different, so
limx->0
1x doesn't exist. (If
both left- and right-limits had been, say, -∞, the limit still wouldn't exist but we could at least use the phrase:
"limit = -∞". This would be the case with
limx->0
-1
x2 = -∞)
(b)
lim
x->3+
x
9 - x2 = -∞ since the numerator has a limit of 3 (a positive number) whereas the
denominator approaches 0 through negative values (and a number near 3
a small negative number is a large negative number).
(c)
lim
x->0+ 2
1/x = ∞ since, as x approaches 0 through positive values, 21/x takes on values which are
"2 raised to a large positive number "... hence the answer "∞".
(d)
lim
x->0- 2
1/x = 0 since, as x approaches 0 through negative values, 21/x takes on values which are
"2 raised to a large negative number" (for example 2-100 = 1
2100 ) hence the answer 0.
We've talked about limits where x approaches a number and y = f(x) becomes infinite. Now let's talk aboutlimits where x becomes infinite and y approaches a number.
We write
limx->∞
x2
2-3x2 = -
13 and we mean that, as x becomes arbitrarily large (and
positive), the values of y = x2
2-3x2 get arbitrarily close to the number -
13 .
Graphically, it means that the graph of y = x2
2-3x2 approaches the line y = -
13 as x
becomes infinite (as in the diagram).
23
Examples:
limx->∞
x
x2+1 = 0
limx->∞
x2
2-3x = - ∞
(meaning that y = x
x2+1 approaches the line y=0) (meaning that y =
x2
2-3x heads south!)
If
limx->∞
f(x) = a or
lim
x->-∞ f(x) = a
then we say that the graph of y = f(x) has a HORIZONTAL ASYMPTOTE, namely: y = a.
CONTINUOUS FUNCTIONS:As x approaches some number, say 5, maybe f(x) has a limit ... and maybe is doesn't. Also, maybe f(x) has avalue at x=5 ... and maybe it doesn't. Even if f(x) had a limit and a value, maybe they're different numbers! Thenicest functions are ones that have a limit and a value and they're the same ... and these functions are calledCONTINUOUS functions.
If (1)
limx->a
f(x) = L and
(2) f(a) exists and
(3)
limx->a
f(x) = f(a)
then f(x) is said to be CONTINUOUS at x = a.PS:S: So why are they the "nicest" functions?P: Because the graph of such a function has no breaks. At every point on the graph of a continuous function there is a left-
limit, a right-limit and a value and they're all the same! The graph could have points or sharp corners, but it has no breaks.However, if a function is discontinuous at a point (say x = 3) then the graph could go in one direction as you approach 3from the left and in another direction as you approach from the right ... and the actual value of the function (hence the pointon the curve at x = 3) might be neither of these!
24
Example: Is f(x) = 4x2-92x-3 continuous at x = 1.5? at x=2?
Solution: Since f(x) doesn't have a value at x = 1.5, it's not continuous there. (It doesn't matter whether it has
a limit!) But, at x = 2, f(2) = 4(22)-92(2)-3 = 7 exists and, as x->2, we have lim
4x2-92x-3 =
lim(4x2-9)lim(2x-3)
= lim(4)lim(x)lim(x)-lim(9)
lim(2)lim(x)-lim(3) = 4(2)(2)- 92(2)- 3 =7 and they're EQUAL, so f(x) is continuous at x=2. (Notice that we didn't
get the limit by substituting x = 2 since that gives the value, not the limit and they may or may not be the same!)
Example:For the function graphed at the right, write limit
statements for :
x->∞, x->-∞, x->0+, x->0-,
x->2-, and x->2+.
Where is the function discontinuous?
Why?
Solution:
limx->-∞
f(x) = -1,
limx->∞
f(x) = -1,
lim
x->0- f(x) = ∞,
lim
x->0+ f(x) = ∞,
lim
x->2- f(x) = 0 and
lim
x->2+ f(x) = 2. The function has NO limit at x = 0 and x = 2, hence is discontinuous there.
(Note that f(x) has a value at x = 2, namely f(2) = 0, but since it has no limit is cannot be continuous there.)
NOW, having established all the "limit terminology" we'll need, stare at the following graph (of a fictitiousfunction f(x)) & verify the comments beneath.
25
LECTURE 3
MORE on LIMITS
TECHNIQUES FOR EVALUATING LIMITS WHEN THE "RULES" DON'T APPLY:In what follows, we consider certain limits to which we cannot apply the various limit rules (for one reason oranother), yet, by using some tricky manipulation we can often put the expression into a form where we CANapply the rules.
• The form ∞∞
limx->∞
x2
2-3x2 =
limx->∞
1
2
x2 -3
= 10-3 = -
13 where we divided numerator and denominator by the largest power of x
(i.e. x2) so that every term then approaches a constant. This is a useful technique when you have
limx->∞
polynomialpolynomial . Note that we can only use the LIMIT RULE
which says lim
f(x)g(x) =
lim f(x)lim g(x) when all
limits exist. In our example, neither limit exists (until we divide numerator and denominator by x2).
• The form ∞ - ∞
limx->∞
( )x+1 - x-1 =
limx->∞
( )x+1 - x-1
x+1 + x-1
x+1 + x-1 where we multiplied both numerator and denominator
by x+1 + x-1 to "rationalize the numerator". We then get
limx->∞
2
x+1 + x-1 = 0 (since the denominator
becomes infinite while the numerator does not). Note that the LIMIT RULE which says lim (f - g) = lim (f) - lim(g) doesn't apply since neither of the limits lim (f) & lim (g) exist.
• The form 00
limx->0
1+x - 1x =
limx->0
1+x - 1
x
1+x + 1
1+x + 1 where we "rationalize the numerator"and
limx->0
1
1+x + 1 =
12
• Reduce the given limit to one you know
Given that
limx->0
sin xx = 1, then we can write
lim
{ }->0 sin{ }{ } = 1 where anything can be plugged into { }. For
example:
lim
2x->0 sin 2x
2x = 1 or, equivalently,
lim
x->-∞ sin 2x
2x = 1 (since x->-∞ will make 2x->0). Of course, in a
real problem, nature isn't so accommodating; you're more likely to see this problem as:
lim
x->-∞ 2-x sin 2x and
it's up to you to change it to the form
lim
{ }->0 sin{ }{ } .
P: This technique of reducing a problem to one you've already solved is very useful, especially for a mathematician. Did I tellyou the story of the mathematician and the engineer?
S: Don't tell me.P: They were both given an empty kettle and asked to make tea. Both did exactly the same thing: fill the kettle with water, boil
the water, put tea bags into the teapot and pour in the boiling water. Then, a new problem. They were given a kettle ofboiling water and asked to make tea. The mathematician first emptied the kettle thereby reducing the problem to one he'dalready solved.
S: I don't like tea.
26
Examples:Evaluate each of the following limits (or explain why the limit doesn’t exist):
(a)
lim
x->5- |x2 - 25|x - 5 (b)
lim
x->5+ |x2 - 25|x - 5 (c)
limx->5
|x2 - 25|x - 5
(d)
limx->0
|2x-1| - |2x+1|
x (e)
limx->∞
2x3 - x + 1
3 - 2x3 (f)
lim
x->π+ cos xtan x
(g)
limx->∞
( )x2+2x - x2-x (h)
limx->∞
2 + xx - 2 (i)
limh->0
x+h - x
h
Remember:
(i) | N | has the value N when N ≥ 0, and the value -N when N < 0. This will enable you to eliminate theabsolute value sign.(ii) The expression x->∞ means x->+∞ (i.e. the "+" is understood).
(iii) The notation “
limx->a
f(x) = ∞ ” means the limit doesn’t exist for a particular reason: the values
of f(x) become arbitrarily large (and positive) for x near "a".Use this notation when appropriate, don't just say "limit doesn't exist".
Solutions:
(a) |x2-25|x 5 =
25-x2
x-5 (since |x2-25| = - (x2-25) when x is close to, but smaller than 5)
= (5-x)(5+x)
x-5 = -(5+x) -> -10.
(b)
lim
x->5+ |x2 - 25|x - 5 =
lim
x->5+ x2-25x-5 =
lim
x->5+ (x-5)(x+5)
x-5 =
lim
x->5+ (x+5) = 10.
(c)
limx->5
|x2 - 25|x - 5 doesn't exist since left- and right-limits differ (as shown above).
(d)
limx->0
|2x-1| - |2x+1|
x =
limx->0
-(2x-1)-(2x+1)
x (since |2x-1|=-(2x-1) when x is close to 0) =
limx->0
(- 4) = - 4.
(e)
limx->∞
2x3 - x + 1
3 - 2x3 =
limx->∞
2 - 1
x2 +
1
x3
3
x3 - 2
= 2-2 = -1 (dividing num. and denom. by x3)
(f)
lim
x->π+ cos xtan x =
lim
x->π+ cos2xsin x (since tan x =
sin xcos x ) = -∞ (since cos
2x->1 and sin x ->0-).
(g) ( )x2+2x - x2-xx2+2x + x2-x
x2+2x + x2-x =
3x
x2+2x + x2-x =
3
1+2/x + 1-1/x ->
32
where, in the last step, we've divided the numerator by x and the denominator by
x2 (which, for positive x, is the same thing as x) .
(h)
limx->∞
2 + xx - 2 =
limx->∞
2/x + 11 - 2/x =
11 = 1.
27
(i)x+h - x
h
x+h + x
x+h + x =
1
x+h + x ->
1
2 x as h->0.
S: I'll tell you something; I can't always tell when something has a limit and when it doesn't. I mean, how should I know whento use some trick if I just looked at the expression and I think it already has a limit ... without doing any work?
P: I don't know what you're talking about. Give me an example.
S: Well, like maybe ... uh, the limit of 1+x - x when x goes to ∞. It's 0 and that's obvious so why would I do some magiclike rationalizing something?
P: That's obvious? Remarkable! It's not obvious to me! I see two huge numbers and when I subtract ...S: You see ∞ - ∞ and that's ... uh, wait ... it isn't 0, is it?
P: No. In fact, 0 happens to be the correct limit, but that's pure luck: 1+x2 - x has a limit of ∞ whereas 1+x - x has alimit of -∞. But let me test you on your ability to recognize expressions where you have to do some work to get the limit ...because it's NOT obvious. In each of the following, x->∞. What's the limit?
(a)
limx->∞
2x
x (b)
limx->∞
x
1 + 1x
(c)
limx->∞
sin 1x
x (d)
limx->∞
10-x2
S: In (a), it looks like ∞∞ so I have to do some work. In (b) it's
∞1 so I have to do some work. In (c) ...
P: No! In (a) you have a huge number divided by a huge number and you can't tell (without doing some work) whether that'ssmall or large. But, in (b), you have a huge number divided by a number which is almost 1. Surely you know what that is.Just think of 1,000,000 divided by 1. It's huge!
S: Okay, so the limit is ∞. In (c) I get ... uh, sin 1∞ in the numerator and that sin 0 and that's 0 so the answer is zero and I didn't
have to do any work. In (d) ...
P: Well ... I wish you wouldn't say 1∞ = 0, it's the limit that's 0, but in (c) you have to look at the denominator too. If the
denominator approached 0 you'd be in trouble because the form 00 isn't one where you can write down the answer. In fact ...
S: But in (c) the denominator isn't approaching 0 ... it's approaching ∞ and that makes the fraction even smaller so it gets to 0
even faster. In (d) I get 10-∞ and I have to do some work. How'm I doin' boss?
P: Terrible! If you don't know anything about 10 raised to a large negative number, try an example! What about 10-1000. Is itlarge? Small?
S: It's ... uh, 1
101000 and I'd say that's pretty small. So I guess
limx->∞
10-x2 = 0, right?
P: Remember this: small numberlarge number
= small number and large numbersmall number
= large number where "large number" could be
large and positive or large and negative, like - 1,000,000.
Examples:
Assuming
limt->0
sin tt = 1, evaluate each of the following:
(a)
limt->∞
t sin 1t (b)
limz->π
sin zz-π (c)
limx->π
(x-π) cot x
Remember:
Given
limt->0
sin tt = 1, we may conclude that
limm->0
sin mm = 1, where m is an expression which approaches zero.
Solutions:
(a)
limt->∞
t sin 1t =
limt->∞
sin 1t
1t
=
limq->0
sin qq = 1 where q =
1t -> 0 as t->∞.
28
(b)
limz->π
sin zz-π =
limz->π
-sin(z-π)
z-π (since sin(z-π) = -sin z) =
limt->0
-sin tt = -1 where t = z-π .
(c)
limx->π
(x-π) cot x =
limx->π
x-πsin x cos x = (-1)(1) using result from (b).
Example:Sketch the graph of a function which satisfies all of the following:
lim
x->-∞ f(x) = -∞,
lim
x->0- f(x) = 0,
lim
x->0+ f(x) = 1,
limx->∞
f(x) = 1
Solution:
The graph of f(x) could look like any of the following:
Examples:A function which has no value at x = a, but has a limit as x->a, may be made continuous (at x = a) by defining
the value to be the limit. For example, f(x) = x2-25x-5 has no value at x = 5, yet
limx->5
f(x) = 10, so the "redefined" function f(x) =
x2-25x-5 for x ≠ 5
10 for x = 5
is continuous
since
limx->5
f(x) = f(5) = 10 .
For each of the following, f(a) is undefined. Determine whether it is possible to define f(a) so as to make thefunction continuous there:
(a) f(x) = sin xx , a = 0 (b) f(x) =
x - 3x - 3 , a = 3
(c) f(x) = e-1/|x| , a = 0 (d) f(x) = |x - 1|x - 1 , a = 1
Solutions:
(a) Define f(0) =
limx->0
sin xx = 1.
(b) Define f(3) =
limx->3
x - 3x - 3 =
limx->3
1
x+ 3 =
1
2 3 .
(c) Define f(0) =
limx->0
e-1/|x| = 0
(d)
lim
x->1-
|x - 1|
x - 1 =
lim
x->1-
-(1-x)
1-x = -1 and
lim
x->1+
x-1
x-1 = 1 differ, so
limx->1
f(x) doesn't exist, so f(x)
29
cannot be made continuous by redefining f(1).
MORE ON LIMITS:PS:S: Since sin x->0 as x->0, then f(x) sin x->0 as x->0, for any function f(x). Right?P: Wrong. If f(x)->∞ as x->0, then the product f(x) sin x could approach anything, depending upon the function f(x). For
example, if f(x) = 1x , then the product approaches 1 and if f(x) =
πsin x
, then the product approaches π .
S: What if f(x) doesn't approach ∞? Then surely f(x) sin x->0 as x->0. Right?P: Sure, if f(x) doesn't get too large.S: Let's see you prove that.P: Okay. Suppose |f(x)| ≤ 1000 for all x. Then |f(x) sin x| ≤ 1000 | sin x |. Further, |f(x) sin x| is an absolute value so it's never
negative, hence 0 ≤ |f(x) sin x| as well. We then have:0 ≤ | f(x) sin x | ≤ 1000 | sin x |
Now we let x->0. Since |f(x) sin x| is stuck between two functions, both of which have a limit of 0 (namely g(x) = 0 andh(x) = 1000 | sin x | ), then |f(x) sin x| also has a limit of 0.
S: Is that some kind of theorem?P: Yes. Here it is:
the SQUEEZE THEOREM
If g(x) ≤ f(x) ≤ h(x) and
limx->a
g(x) =
limx->a
h(x) = L, then
limx->a
f(x) = L as well.
In words: If f(x) lies between two other functions, and they have the same limit, then f(x) also has this limit.
This theorem is often used to show that some limit of the form
limx->a
f(x) g(x) is 0, when one factor approaches 0
(say, lim (f) = 0) but the other factor doesn't have a limit ... but is bounded. In that case you can't use the LIMITRULE which says: lim (f g) = lim (f) lim (g) because this RULE requires that both limits, lim (f) and lim (g),exist. (It would be nice if you could use the RULE; you'd say lim (f g) = lim (f) lim (g) = 0 lim (g) = 0, but ifthere is no number "lim(g)", then what's the meaning of "0 lim (g) = 0"? It's like multiplying zero by a yellowrose. Is it zero?)
Example: Evaluate
limx->0
x sin 1x . (Note that sin
1x doesn't have a limit ... and don't confuse this problem
with
limx->∞
x sin 1x which can be written
limx->∞
sin 1x
1x
= 1 which is the same as
limm->0
sin mm
with m = 1x ,
hence its limit is 1.)
Solution: Since | sin 1x | ≤ 1, then 0 ≤ |x sin
1x | ≤ | x | and since both sides of this inequality have the same
limit, namely 0, then |x sin 1x |->0 as well. In this example, it's tempting to write: -1 ≤ sin
1x ≤ 1 so that
-x ≤ x sin 1x ≤ x and then use the SQUEEZE theorem. Unfortunately, x sin
1x ≤ x isn't even true for negative x.
Try x = -π for example. We'd get -π sin (-π) ≤ -π which says 0 ≤ -π (which certainly isn't true). Moral? If you'dlike to prove that some limit is zero, you can prove that its absolute value has a limit of zero.
In fact, while we're at it, let's say something about y = f(x) sin 1x . Since sin
1x oscillates between -1 and +1, then
y = f(x) sin 1x oscillates between y = - f(x) and y = f(x) and, in fact, touches the curve y = f(x) whenever sin
1x
has the value +1 and it touches y = - f(x) whenever sin 1x = -1.
30
The same is true of the graph of y = f(x) sin x (or, forthat matter, y = f(x) cos x). The graph oscillatesbetween y = f(x) and y = - f(x). It's even moreinteresting if f(x) changes slowly and sin x changesrapidly. We could accomplish this by considering
f(t) sin wt where w is a very large number (hencesin wt oscillates rapidly). Then f(t) sin wt oscillates so
rapidly between y = f(t) and y = - f(t) that it almostfills the space between. Indeed, if f(t) is the voltageproduced by a microphone while recordingBeethoven's Fifth
Symphony (and that's why I changed the name of thevariable to t = time!), and if we modify theamplitude of a second sinusoidal voltage, sin wt, soit becomes
f(t) sin wt, we've got amplitude modulation (AM forshort) and we can send it up a big antenna and youcan receive it on your car antenna and your AMradio can delete the sin wt and recover the f(t) (i.e.demodulate the radio signal) and send f(t) to yourspeaker and you can relax with Beethoven. (In fact,each radio station has its own distinct w which youcan "tune in".)
PS:
S: If I have to evaluate, say,
limx->1
3x
(x-1)2 , can't I just say it's
30 = ∞?
P: If we agree that the expression 30 is shorthand for the limit of a ratio where the numerator has a limit of 3 and the
denominator has a limit of 0, then it's okay ... sort of. However, most profs get very nervous when students use this notation.They're likely to cross-multiply and get 3 = 0 x ∞ which is meaningless. Besides (and this is important), the denominator
could approach 0 through negative values, such as would be the case for
limx->1
3x
-(x-1)2 where the limit is -∞ (and you'd
be tempted to still write 30 = ∞, and get the wrong answer). Also, for the case
limx->1
3xx-1
, you'd still get 30 but now the left-
limit is
lim
x->1- 3xx-1
= -∞ whereas the right-limit is
lim
x->1+ 3xx-1
= ∞. So you have to be very careful.
S: Can I write
lim
x->1- ( )
3xx-1
= 3-0 = -∞ and
lim
x->1+ 3xx-1
= 3+0
= ∞ where I show the "sign of zero" ... or at least how I
approach 0 ... so I get the right sign?P: Sure, as long as you indicate the convention you're using so everybody who reads what you write will
understand what you're saying. For example, it's often convenient to write
lim
x->1- 3x(4-5x2)
x-1 =
(+)(-)(-0)
= ∞ (since the two negatives give a positive result).
Using this shorthand notation indicates that the graph of y = 3x(4-5x2)
x-1 has a vertical asymptote and
that the curve goes due north as x approaches 1 from the left.
S: So now that we have all this stuff about limits, what good is it?P: The good part is yet to come. Calculus comes in two flavours: differential calculus and integral calculus. Differential
calculus is about rates of change and the slope of tangent lines to curves ... the DERIVATIVE. Integral calculus is about
31
areas under curves and breaking down a complicated problem into simpler ones and summing ... the DEFINITEINTEGRAL. Both the derivative and the integral are defined in terms of a limit, hence we began our study with LIMITS (...unfortunately, it's perhaps the most difficult part of our study!). Next we'll turn to the differential calculus, and theDERIVATIVE ... but first a few problems.
Problems:1. Evaluate each of the following (or explain why the limit doesn’t exist):
(a)
limx->4
|x2-16|x-4 (b)
limx->0
|2x2-1| - |2x2+1|
x2
(c)
limx->∞
4x3 + x - 1
4 - 2x3 (d)
limx->∞
( )x2 - 5x - x2 + x
2. Determine the points of discontinuity (if any) for the function f(x) =
x
2 x>0 1 x=0
-x2 x<0
Solutions:
1. (a) For x > 4, |x2-16|x-4 =
(x-4)(x+4)x-4 = x+4->8 as x->4+
For x < 4, |x2-16|x-4 =
-(x-4)(x+4)x-4 = - (x+4)->-8 as x->4-
Hence
limx->4
|x2-16|x-4 doesn’t exist (... left and right limits don’t agree).
(b)
limx->0
|2x2-1| - |2x2+1|
x2 =
limx->0
- (2x2-1) - (2x2+1)
x2 =
limx->0
(-4) = -4
where |2x2-1| = -(2x2-1) since 2x2-1 < 0 when x is near 0 (and, of course, 2x2+1 > 0).
(c)
limx->∞
4x3 + x - 1
4 - 2x3 =
limx->∞
4+1/x2-1/x3
4/x3-2 =
4-2 = -2.
(d) ( )x2-5x - x2+xx2-5x + x2+x
x2-5x + x2+x =
-6x
x2-5x + x2+x =
-6
1-5/x + 1+1/x ->-3 as x->∞.
2. Since f(x) is defined differently for x > 0 and x < 0, we consider both left- and right-limits:
lim
x->0- f(x) =
lim
x->0- (-x2) = 0 and
lim
x->0+ f(x) =
lim
x->0+ (x2) = 0
Since they're the same, this function does have a limit:
limx->0
f(x) = 0. However, it's not equal to f(0) = 1, so the
function is NOT continuous at x = 0. For every other value of x, say x = a > 0, we would get
limx->a
f(x) =
limx->a
x2 =
limx->a
x
limx->a
x = (a) (a) = a2, using a LIMIT RULE. Since this is equal to f(a) = a2 (for a
> 0) , the function is continuous for every x > 0. Similarly we could prove it continuous for every x = a < 0.Hence x = 0 is the only discontinuity.
Comments: If we've just found a limit, say
limx->0
sin xx = 1, then we've actually shown that sin x and x have
very nearly the same value when x is small (since their ratio is nearly "1"). For example, choosing x = .0123 wefind that sin (.0123) = .01229969 (and if we use a calculator for this evaluation, we make sure it's in radian
mode, else sin x and x are NOT close in value for small "x"). In fact,
limx->0
sin xx = 1 says that the two curves, y
32
= sin x and y = x, are very nearly coincident for x small. It may not be such a surprise, then, to learn that y = x isthe tangent line to the curve y = sin x, at x = 0 ... which brings us to tangent lines.
LECTURE 4
the DERIVATIVE and its RULES
the DERIVATIVE:Our problem is to determine the slope of thetangent line to the curve y = f(x), at x = a. Todo this we consider two points on the curve;one at x = a (and y = f(a)) and a second point atx = a + h (and y = f(a+h)). The slope of the line
joining these two points is f(a+h) - f(a)
h . On
the other hand, we could take the second pointas x (and y = f(x)) in which case the slope ofthe line joining the two points would bef(x) - f(a)
x - a . Regardless of what we call the
second point, the slope of the tangent line at x= a will be obtained by taking the limit of thisslope as h->0, or as x->a.
f '(a) =
limh->0
f(a+h) - f(a)
h =
limx->a
f(x) - f(a)x - a
This limit is called the derivative of f(x) with respect to x, at the place x = a.
Example: Determine the derivative of f(x) = x at x = 2.
Solution: f '(2) =
limh->0
f(2+h) - f(2)
h =
limh->0
2+h - 2
h =
limh->0
2+h - 2
h
2+h + 2
2+h + 2
=
limh->0
1
2+h + 2 =
1
2 2 (where we have "rationalized the numerator") .
We could also use the second form for the derivative:
f '(2) =
limx->2
f(x) - f(2)
x - 2 =
limx->2
x - 2x - 2 =
limx->2
x - 2
x - 2
x + 2
x + 2 =
limx->2
1
x + 2 =
1
2 2 .
Of course, we could also use
limt->2
f(t) - f(2)
t - 2 or
limu->0
f(2+u) - f(2)
u since it matters little what names we give to the
variables. If you understand the diagram, and the fact that we take one point as x = a and some second point(giving it any name), and let the second point approach the first, then the limiting value of the slope of the linejoining the two points is the derivative of f(x) at x = a (that is, f '(a)).
If the limit which defines f '(a) exists, we say f(x) is differentiable at x = a.
The above limit definition (take your pick which one) gives the derivative of f(x) at some point "a". If we want
33
the derivative at a variable point "x", we just replace "a" with "x" in the definition:
f '(x) =
limh->0
f(x+h) - f(x)
h =
limt->x
f(t) - f(x)
t - x
S: That's confusing, I mean ...P: Pay attention. Suppose we want to compute the derivative of f(x) at, say, x = 6. Okay, (6, f(6)) is the point in question. Now
pick another point.S: Okay, I pick x = 7.P: Sorry, I mean pick a name for another point. We have to let this second point approach the first and we can't do that if it's
stuck at 7. See? Okay, pick a name.S: Sam.P: Hmmm. The point with x = Sam has y = f(Sam). Now what's the slope between these two points on y = f(x)?
S: It's ... uh, f(Sam) - f(6)Sam - 6
, right?
P: Good! Now let Sam->6 and you've got ... what?S: Huh? I don't understand the question.
P: You've got
lim
Sam->6 f(Sam) - f(6)Sam - 6
and that's f ' (6), the derivative of f(x) at x = 6. Nice, eh? And now you can see that it
makes little difference what we call the second point so long as it's approaching the first. I'll give you yet another notation,maybe more appealing.
The change (or increment) in x, which we call "h" above, is sometimes called Dx ( a convenient notation whichindicates that it's a small increment if the variable "x"). The corresponding change in y, namely
f(x+Dx) - f(x), is often denoted by Dy. This makes the definition of the derivative take the form:
dydx =
lim
Dx->0 DyDx =
lim
Dx->0 f(x+Dx) - f(x)
Dx
which is familiar to many and is the reason for the notation dydx for the derivative of y with respect to x (i.e.
DyDx
becomes dydx , in the limit)* . If y = f(x) we use
dydx or y' or f '(x) or sometimes just f ' to represent the
derivative.
Using the limit-definition of the derivative we can show that ddx x
n = n xn-1 where n is any integer. We illustrate
with n = 5.
If f(x) = x5, then f(t) - f(x)
t - x = t5 - x5
t - x = (t - x)(t4+t3x+t2x2+tx3+x4)
t - x
= t4 + t3x + t2x2 + tx3 + x4 -> x4 + x4 + x4 + x4 + x4 = 5 x4 as t->x
In fact, ddx x
p = p xp-1 works for any exponent, whether integer or not.
For example, ddx x =
ddx x
1/2 = (1/2) x-1/2 = 1
2 x .
the DIFFERENTIAL
We've already mentioned that, after having evaluated some limit such as
limx->0
sin xx = 1, we can interpret this to
* This notation is due to Gottfried Wilhelm Leibniz (1646-1716), who created calculus ... along with Isaac Newton(1642-1727), who discovered the ideas independently. Newton called derivatives "fluxions" and used a notationmore like x' (which is still used today). In 1924 a letter written by Newton was discovered in which Newtonacknowledged that some of his early ideas came from Fermat (1601-1665) who found a method for constructingtangents to curves and maxima and minima.
34
mean that sin x and x are very close in value when x is small. The same holds for the limits which yield thederivative. Let's see how that works:
We have y = x3 and have just computed
lim
Dx->0 DyDx = 3x
2 , so we know that DyDx is very nearly 3x
2 when Dx is
small. What does that mean? It means that when x changes from x to x+Dx, the corresponding change in y = x3,
namely Dy = (x+Dx)3 - x3, is such that DyDx ≈ 3x
2 and that means we'd expect Dy ≈ 3x2 Dx .
Is that useful? Yes. If x is the length of the side of a cubical box, then y = x3 is its volume. Now suppose the box
is to be built with x = 2 metres (so the volume is 23 = 8 metres3) and there are small errors in building the boxand the sides are made to within .01 metres . What about the volume? We put x = 2 and Dx = .01 (because this
is the possible error, or change, in x) and get an estimate of the change in volume, namely Dy ≈ 3x2 Dx = 3 (2)2
(.01) = .12 metres3.
S: I don't need calculus for that! I'd just say (2.01)3 - 23 and that's the change in volume, and that's .121 m3 and that'd be exact... no estimate, no approximation.
P: Of course I was giving you a simple problem, just to illustrate the idea of using the derivative to estimate small changes. Icould give you a tougher problem. Here it is:
Example: When x = 49, y = x has the value 7. Estimate the change in y when x changes to 47.
Solution: For y = x1/2, then dydx =
12 x
-1/2 = 1
2 x so we can estimate changes in y using
DyDx ≈
1
2 x or
Dy ≈ 1
2 x Dx and if we put x = 49 and Dx = -2 (because x changes from 49 to 47, a change of -2), we get
Dy ≈ 1
2 49 (-2) = -
17 . Conclusion? 47 is approximately: 7 -
17 ≈ 6.857
In general we have the following prescription for estimating changes in the value of a function when the variablechanges by some small amount:
We have y = f(x) and have just computed
lim
Dx->0 DyDx = f '(x), so we know that
DyDx is very nearly f '(x) when Dx
is small. What does that mean? That means that when x changes from x to x+Dx, the corresponding change in y
= f(x), namely Dy = f(x+Dx) - f(x), is such that DyDx ≈ f '(x) and that means we'd expect Dy ≈ f '(x) Dx .
Dy ≈ f'(x) Dx is called the DIFFERENTIAL of y = f(x),and it depends upon x and upon Dx, the DIFFERENTIAL of x,
and it gives an estimate of the change in y when x changes by Dx.
Example: The distance travelled after a time t hours is x(t) = 100 + t3 metres. Estimate the distance travelledfrom t = 2 hours to t = 2.1 hours.
Solution: Since DxDt ≈
dxdt = 3t
2, we compute the "differential of x", namely Dx = dxdt Dt = 3t
2 Dt.
Substituting t = 2 and Dt = .1 we get Dx = 3 (2)2(.1) = 1.2 metres as our estimate of the distance travelled.
P: Do you see anything interesting here?S: Nope.
P: If I told you that the velocity was dxdt = 3t2 = 3(2)2 = 12 m/hr at the end of 2 hours, then I asked how far would you would
travel in the next tenth of an hour, what would you say?S: Uh ... (12)(.1) = 1.2 metres.
35
P: Right, and that's just what the differential is doing for us. It gives quite a reasonable approximation if you don't changethings too much.
S: But .12 metres is exact, isn't it?P: No, because the velocity is changing, even during the time from t = 2 to t = 2.1 hours. In fact, the velocity is increasing so
the distance would actually be somewhat larger. To be exact, the distance would be (100 + (2.1)3) - (100 + 23) = 1.261metres.
more on DERIVATIVES
S: I notice that you keep saying "the derivative with respect to x". Why the "with respect to"?
P: If y depends upon x and x changes by Dx and the corresponding change in y is Dy, then the limit of DyDx
is the derivative of
y with respect to x. Now if y depended upon somebody called u, and u changed by Du, then
limDu->0
DyDu
would be ...
S: Don't tell me ... the derivative of y with respect to u.
P: Right! And one other thing. If y is measured in hectares and x is measured in degrees Celsius, then
lim
Dx->0 DyDx
= dydx
is
measured in hectares per degree Celsius. See? And if u were measured in volts, then
lim
Du->0 DyDu
= dydu
would be measured
in hectares per volt. See? The "with respect to" is important ... although we'll often omit this phrase if it's obvious who we'redifferentiating with respect to. See?
S: I guess so.P: And one more thing. Since the derivative is a "rate of change", it tells how rapidly y changes when x changes. That is, if
dydx
= 10, say, it means that y is increasing 10 times more rapidly than x is. If dydx
= - 1/3 then it means that y is decreasing
as x increases (because dydx
is negative) and it's decreasing 1/3 as rapidly as x increases ... and ...
S: Yeah, I got it.P: And one more thing. Don't get tied to any particular set of labels. The independent variable could be called y or z or V or
even x, and we could use the notations dydx
or dzdp
or dVdr
or dxdt . See?
S: Can we keep going?P: Not yet -- we should recognize when things increase and decrease and how the derivative gives us this info and ...S: A picture is worth a thousand words, remember?P: Okay, here's the graph of some invented function f(x).
For a while it increases, then it decreases, then itincreases again and so on. These increases and decreasesare reflected in the value of the derivative, f ' (x), so if weplot f ' (x) versus x we'll find that when f(x) is increasing,f ' (x) will be positive and when f(x) is decreasing f ' (x)will be negative. See how it goes?
S: Yeah ... and it looks like ... uh ... what's happening at x =.8 and x = 1.5 and x = 2.2 or thereabouts?
P: When f(x) stops going up and starts coming down, it hasa horizontal tangent so its derivative f ' (x) is zero. Thathappens at about x = .8 and again at about x = 2.2 and atx = 1.5 the derivative f ' (x) is again zero because f(x)again has a horizonatl tangent.
S: Hmmm ... does this kind of thing ever show up in realproblems?
P: Yes ... trust me.
36
DIFFERENTIATION RULES
If f(x) and g(x) are both differentiable at x, then:
SUM: ddx (f(x) + g(x)) = f '(x) + g'(x) DIFFERENCE:
ddx (f(x) - g(x)) = f '(x) - g'(x)
PRODUCT: ddx ( )f(x) g(x) = f(x) g'(x) + f '(x) g(x)
QUOTIENT: ddx
f(x)g(x) =
g(x) f '(x) - f(x) g'(x)
g2(x)
with the proviso that, in the QUOTIENT rule, g(x) ≠ 0 (else the expression in undefined!).
Examples:
• Suppose we knew that ddx sin x = cos x and
ddx cos x = - sin x. Then
ddx tan x =
ddx
sin xcos x =
cos x (cos x) - sin x (-sin x)
cos2x =
cos2x + sin2x
cos2x =
1
cos2x = sec2x
(using the QUOTIENT rule).
• We can also use the PRODUCT rule to compute: ddx x
3x5 = x3(5x4) + (3x2)x5 = 8x7 which (fortunately)
agrees with ddx x
n = n xn-1 when n = 8.
the CHAIN RULE:One of the most useful rules (all of which can be derived from the definition of the derivative!) is:
the CHAIN RULE
If y = f(u) and u = g(x), then:dydx =
dydu
dudx
ordydx = f '(u) g'(x)
Examples:
• If y = sin x3, we can write this relation as y = sin u and u = x3 so thatdydx =
dydu
dudx =
ddu sin u
d
dx x3 = (cos u) (3x2) = 3x2 cos x2. Hence
ddx sin x
3 = 3x2 cos x2.
• Suppose that y = sin( )cos x3 . We can write y = sin u, u = v , v = cos w and w = x3. Then
dydx =
dydu
dudv
dvdw
dwdx and all are easy differentiations! We get:
dydx = ( )cos u
1
2 v( )-sin w ( )3x2 = -
3x2 cos( )cos x3 sin x3
2 cos x3 .
Example: Find the equation of the tangent line to the curve y = 2x sin x at the point (π2 ,π).
Solution:dydx = 2x (cos x) + 2sin x (using the PRODUCT rule) is the slope of the tangent line at any place x.
37
For x = π2 we get
dydx = 2
π2 cos
π2 + 2 sin
π2 = 0 + 2 = 2. We use the point-slope form of the equation of a
straight line to get: y - π
x - π2
= 2 or y = 2x as the tangent line. (Note that this tangent line passes through the
origin. Try this problem: "Find a point on y = 2x sin x such that the tangent line passes through the origin."Pretend you don't know the answer!)
Example: Find the normal line to the curve y = 2x sin x at the point (π2 ,π).
Solution: Since the tangent line has slope 2, the normal line (which is perpendicular to the tangent line) has
slope - 12 (the product of the slopes must be -1) and passes through the same point (
π2 ,π). Its equation is then:
y - π
x - π2
= - 12 or x + 2y =
5π2 (and we make a quick check that (
π2 , π) does lie on the line) .
HIGHER DERIVATIVES:Starting with some differentiable function y = f(x) we generate a second function, f '(x), by differentiating:dydx = f '(x) which, as we've said, can be interpreted as the slope of the tangent line to the curve y = f(x) at the
place x. We can also differentiate again to obtain the second derivative: ddx
dy
dx = ddx f '(x) which is written
using the notation: d2y
dx2 = f ' '(x). If f '(x) is increasing, then
its derivative will be positive, i.e. d2y
dx2 > 0. If f '(x)
is decreasing, then d2y
dx2 < 0. Geometrically
speaking, if d2y
dx2 > 0 then the tangent line has
an increasing slope so the curve is concave up.
If d2y
dx2 < 0 then the tangent line has an
decreasing slope so the curve is concave down.
38
For example, if s is the distance travelled after a time of
t hours (measured in kilometres, say), then v = dsdt
is the velocity (in kilometres/hour ) so v > 0 when sis increasing and v < 0 when s is decreasing and soon. In addition, the second derivative of s, the
acceleration a = dvdt =
Error!, tells when v is increasing or decreasing sowhen a > 0 the velocity is increasing and so on.
To the left are three graphs. One of them is s(t), thedistance travelled (plotted versus t), one of them is
the velocity v(t) = dsdt and one is the acceleration, a
= d2s
dt2 .
Which is which?
P: Are you listening? Which is which?S: I'd say that ... uh, one is the slope of the other and ... uh
... I don't understand the question.P: The middle one is s(t) versus t and its slope is given by
the first graph and you'll notice that at the point markedwith a big dot the distance is constant for awhile so thevelocity is zero. Further, the first graph decreases thenincreases so its derivative must first be negative thenpositive and that's what happens in the last graph so that
must be dvdt , the acceleration.
Finally, you may notice that, since a = d2s
dt2 is the 2nd derivative of s(t) it gives the concavity of s(t), so when the middle
graph is concave down, the acceleration is negative. Nice, eh?S: Why would anybody be interested in the slope of the tangent line to some curve, or, for that matter, whether it's concave up
or down? I can't get too excited about acceleration. Is that why we're studying calculus?P: No. Calculus involves derivatives and these occur everywhere ... the universe unfolds according to equations involving
derivatives ... and that's why we study derivatives.S: For example?
P: We've already mentioned that if s is the distance travelled by an object, then v = dsdt is its velocity and
dvdt =
d2s
dt2 is its
acceleration. Newton's law of universal gravitation looks like d2r
dt2 = -
k
r2 , again involving derivatives. The spot which
paints the picture on your TV screen moves across the screen according to d2s
dt2 + k(s2-1)
dsdt + s = 0. See? Derivatives! A
hot object at temperature T cools according to dTdt = - k (T - c). More derivatives. The volume of a sphere is V =
4π3 r3
and the surface area is dVdr
. If C(x) is the cost of producing x items then dCdx
is the marginal cost per item and if R(x) is the
revenue then dRdx
is the marginal revenue. The consumer price index ...
S: zzzzz
39
Example: A lake is stocked with 1000 fish. It is found that N(t), the number of fish after t years, increases so
that its rate of increase is governed by the equation: dNdt = k N (10,000 - N) (called the "Logistic Equation").
Show that the rate of change, dNdt , is a maximum when the population is 5,000 fish.
Solution: We determine how rapidly dNdt increases by taking its derivative:
d2N
dt2 =
ddt
dNdt = k N (-
dNdt ) + k
dNdt (10,000 - N) where we've differentiated k N (10,000 - N) as a product. Hence
d2N
dt2 = k
dNdt (10,000 - 2N) . Note, first, that
dNdt = k N (10,000 - N) is positive when 0 < N < 10,000. Now, when N
< 5,000 we have d2N
dt2 > 0 so
dNdt is increasing. When N > 5,000 we have
d2N
dt2 < 0 so
dNdt is decreasing. Hence
the maximum rate of increase occurs when a population of 5,000 is reached.
Note: If the fish population obeyed the Logistic Equation, dNdt = k N (10,000 - N), then N(t) increases until
N = 10,000 which is the eventual population of the lake (called the "carrying capacity" of the lake). Normally one
wouldn't know the eventual population so we'd have to assume an equation dNdt = k N ( P - N) where the
eventual population is some unknown constant P. Then we'd drop 1000 fish in the lake and each year we'd
measure the population, N(t). The rate of increase dNdt will increase at first, then decrease. When this happens
the population is 12 the carrying capacity. If the population at that time is, say, 15,000 then we can expect the
eventual fish population to be 30,000.
(A nice problem: Assuming dNdt = k N ( P - N), where k and P are constants, show that the rate of change,
dNdt , is a
maximum when the population is P2 fish.)
Example: An object moves in a straight line so that s, its distance (in metres) after a time t (seconds), is givenby the relation: s = t sin(πt). Determine its velocity and acceleration at time t = 1.
Solution: We need to find the velocity dsdt and acceleration
d2s
dt2 . First,
dsdt = πt cos(πt) + sin(πt) using the
PRODUCT rule, then d2s
dt2 = - π2t sin(πt) +2π cos(πt) and, at t = 1, we get
dsdt = - π and
d2s
dt2 = - 2π.
(Note that dsdt < 0 means that s is decreasing; the object is getting closer. Also,
d2s
dt2 < 0 means that
dsdt is decreasing;
the object is slowing down.)
LECTURE 5
IMPLICIT DIFFERENTIATION & TRANSCENDENTAL FUNCTIONSSome Notes:• If f '(x) = 0 over some interval, then f(x) = constant over that interval (i.e. the graph of y = f(x) is a horizontal
line). This means that y is NOT changing when x changes.• Note the interpretation of the derivative as a rate of change.
If
lim
∆x->0 ∆y∆x = 6 then y changes 6 times more rapidly than x
40
This interpretation was used by Newton*, who imagined x moving along a line at constant speed, and y = f(x)
moving along another, parallel line. Then the derivative dydx indicated how much faster (or slower) y was
moving. In fact, it's this interpretation of the derivative which makes the differential calculus such an importanttool (and not the interpretation as the slope of the tangent line to a curve ... as if anybody were interested in theslope!).
• If y = f(u) depends upon u and u = g(x) depends upon x then
lim
∆u->0 ∆y∆u tells how rapidly y changes when u
changes ("y runs 6 times faster than u"), and
lim
∆x->0 ∆u∆x tells how rapidly u changes with x ("u runs 7 times
faster than x"), and
lim
∆x->0 ∆y∆u
∆u∆x then gives how rapidly y changes with x ... "y runs (6)(7) = 42 times faster
than x" ... and now the CHAIN rule seems almost trivial!
IMPLICIT DIFFERENTIATION:Sometimes we're given a relation between x and y which, although is does define y as a function of x, doesn't
give y explicitly in terms of x. For example, consider the relation y3 + y5 = x. For x = 1 the y-value must be
found from y3 + y5 = 1 and there is only one y-value which satisfies this equation. (Can you prove this?)Similarly, for any other value of x there will be a single value of y which satisfies the equation. (Can you provethis?) Consequently, y is a function of x (even though we can't usually find the y-value ... except for certainpeculiar x-values such as x = 0 in which case y = 0, or x = 2 in which case y = 1). However, remarkably, we can
find dydx as follows:
Regarding y as a function of x we differentiate the relation y3 + y5 = x with respect to x and getddx ( )y3 + y5 =
ddx x or, using the CHAIN rule we get
ddy ( )y3 + y5
dydx = 1 hence (3y
2 + 5y4) dydx = 1,
so dydx =
1
3y2 + 5y4 . If we know a point on the curve, such as (2,1), then the slope of the tangent line at that point is
dydx =
1
3(12) + 5(12) =
18 .
Example: Find dydx at the point (1,1) if xy = sin
πy2 .
Solution: First we check that (1,1) does lie on the curve! It does.
Next we differentiate the entire equation, regarding y as a function of x: ddx (xy) =
ddx sin
πy2 .
This gives: x dydx + y =
π2 cos
πy2
dydx . Now substitute the given point, x = 1, y = 1 and get:
dydx + 1 = 0 so
dydx = -1. (Note: the tangent line would be
y - 1x - 1 = -1 or y = -x + 2)
Example: Assuming ddx x
n = n xn-1 holds for integers n, prove that it holds for fractions pq as well.
Solution: Let y = xp/q, then yq = xp and we differentiate implicitly: ddx y
q = ddx x
p gives
q yq-1 dydx = p x
p-1 (where we need only use the differentiation formula for integers p and q). Then
* Isaac Newton (1642-1727) was one of the greatest geniuses of all time and, along with Archimedes and Gauss, oneof the top three mathematicians. He discovered, among other things, the law of universal gravitation, the binomialtheorem, the breaking of white light into its constituent colours and, of course, the differential and integral calculus... and he did most of this in his twenties! (How does that make you feel?)
41
dydx =
p xp-1
q yq-1 =
pq
xp-1
xp(q-1)/q (where we have substituted y = xp/q) =
pq x
p/q - 1 ... and the formula works for any
rational number!PS:S: What about the rest of the numbers, like 2 and π which aren't rational numbers?
Is ddx
xπ still equal to π xπ-1 ?
P: Yes ... but it isn't easy to prove.
S: One other thing. You said that the standard equation of a circle, x2 + y2 = a2 (where "a" is a constant), doesn't define y as afunction of x ... since there are usually two y-values for each x-value. Right?
P: Right.
S: Then if I go right ahead and take ddx
of everything, that is, I differentiate implicitly, even though y is NOT a function, I get
2x + 2y dydx
= 0 so dydx
= - xy . Is that wrong?
P: Actually, it's okay ... sort of. Since you can extract a function from this relation, y = a2 - x2 for example, thendydx
= - xy gives the slope of the tangent line at a point (x,y) on the graph of this extracted function. See?
S: No.
P: Well, let's take y = a2 - x2 for example. Then dydx
= ddx
( )a2 - x2 1/2
= 12 ( )a2 - x2
-1/2(-2x) = -
x
a2 - x2 which, of
course, is - xy . If we take another function out of x2 + y2 = a2, say y = - a2 - x2 , then
dydx
= x
a2- x2 = -
xy again.
S: Are they the only two functions you can "extract" from x2 + y2 = a2?
P: No. We can also define y = a2 - x2 when -a ≤ x ≤ 0 and
y = - a2 - x2 when 0< x≤ a. This is a function, with a single definedvalue for each x in the domain -a ≤ x ≤ a; its graph is shown in thediagram. (By the way, it doesn't have a derivative at x = 0).
Whether we compute dydx
for negative x (using y = a2 - x2 ) or for
positive x (using y = - a2 - x2 ) we'd get the result dydx
= - xy .
So the dydx
we get from implicit differentiation actually gives us the
correct slope no matter what function we extract from the relation
(provided dydx
exists!). That's because dydx
= - xy doesn't give the slope
at some x-value (where there may be two y-values satisfying therelation), it gives the slope at a point (x,y).
S: Wait a minute. You said it doesn't have a derivative at x = 0? Can youprove that?
P: Remember the definition! In order to have a derivative, the limit
limx->0
f(x)-f(0)x-0
must exist.
42
In particular the right-hand limit must exist and that's:
lim
x->0+ - a2-x2 - a
x =
-2a+0
= -∞.
In fact, it's clear from the graph that as x approaches 0 from the right, the chordjoining the two points (one being the point (0,a)) becomes vertical and its slopebecomes large and negative, approaching -∞.
S: Do I have to know this for the final exam?P: No.
Examples:For each of the following:
sketch a graph of the function indicating asymptotes and discontinuities, andwrite "limit statements" for x->-∞, x->∞, and
if f(x) has a discontinuity at x = a, write two limit statements for x->a- and x->a+.
(a) f(x) = 4
x2+2x (b) f(x) =
4 - x2
x2+2x (c) f(x) =
sin xx
Solutions:
(a) f(x) = 4
x2+2x =
4x(x+2) has discontinuities at x = 0 and x = - 2
lim
x->-∞ f(x) = 0,
limx->∞
f(x) = 0,
lim
x->0- f(x) =
4(-0)(2) = -∞,
lim
x->0+ f(x) = ∞,
lim
x->-2- f(x) = ∞ and
lim
x->-2+ f(x) = -∞
(b) f(x) = 4-x2
x2+2x =
(2-x)(2+x)x(x+2) =
2-xx (for x ≠ - 2)
has discontinuities at x = 0 and x = - 2.
Since f(x) =
4
x2 - 1
1 + 2x
then
lim
x->-∞ f(x) = -1,
limx->∞
f(x) = -1
lim
x->0- f(x) = -∞,
lim
x->0+ f(x) = ∞,
lim
x->-2- f(x) = -2,
lim
x->-2+ f(x) = -2
(c) f(x) = sin xx has a
discontinuity at x = 0.
limx->-∞
f(x) = 0 and
limx->∞
f(x) = 0
43
(Note: 0 ≤ | f(x) | ≤ 1|x| , then we use the SQUEEZE theorem to get the limit.)
limx->0-
f(x) =
lim
x->0+ f(x) = 1
P: Notice anything interesting about the graph of y = 1x sin x ?
S: Nope.
P: It oscillates between y = 1x and y = -
1x . Remember? Since sin x oscillates between +1 and -1, then y = f(x) sin x oscillates
between y = f(x) and y = - f(x) and in this case ...S: Yeah, I remember.
Examples:The greatest integer function [[x]] is defined as "the greatest integer not exceeding x".(For example, [[-5.7]] = -6, [[-5]] = -5, [[-4.8]] = -5, [[4.8]] = 4 )
(a) Sketch a graph of f(x) = [[x]] for -4 ≤ x ≤ 4 and identify points of discontinuity.
(b) Sketch a graph of f(x) = [[x + 12 ]] for -4 ≤ x ≤ 4 and identify points of discontinuity (and note that f(x)
rounds any number "x" to the nearest integer!)
Solutions:
(a) For f(x) = [[x]], points of discontinuity occur at every integer.
(b) For g(x) = [[x + 12 ]], points of discontinuity occur halfway between successive integers.
P: Remember the sign in the post office ... about the cost of postage? It was C(x) = 10 if 0<x<1 15 if 1≤x<2 20 if 2≤x<3
.
S: Nope.P: Well, look at your notes. Anyway, the cost could also be written C(x) = 5 + 5[[x+1]]. Do you see that?S: Nope.P: Can you imagine such a sign ... hanging in a post office?S: Nope.P: Your cerebral prowess leaves something to be desired.
44
TRANSCENDENTAL FUNCTIONS: trig, exponential and log functions
Many functions we deal with can be evaluated for every x (in the domain of the function) using a $4.95calculator (that can only add, subtract, multiply, divide and extract roots). These are called algebraic functions.
(e.g. x2+1 , x
x1/3+x and a+bx+cx2+dx3.) Any function which is not algebraic is called transcendental. These
include sin x, tan x (and the other four trig functions) and 10x (and all the exponential functions ... with anybase), log2 x (and all log functions, with any base) and many others. Unfortunately, the RULES for
differentiation (product, quotient, Chain Rule, etc.) are of little help if we wish to differentiate a transcendentalfunction ... unless, of course, we can express a transcendental function in terms of algebraic functions, usingproducts quotients or composite functions .... but then they wouldn't be transcendental! Hence we must resort tothe definition of the derivative ... and since the derivative is defined in terms of a limit, we can expect to findsome weird limits which require some ingenuity to evaluate.
• To deduce ddx sin x = cos x requires evaluating the limit:
limh->0
sin (x+h) - sin x
h =
limh->0
2 cos (x+h2 ) sin
h2
h (using a magic trig identity)
= cos x
limh->0
sin h2
h2
, hence reduces to the evaluation of a weird limit:
limt->0
sin t
t = 1.
(We won't prove this, but we'll use it ... as we've already done.)
the TRIG FUNCTIONS and their derivatives
Beginning with ddx sin x = cos x we can generate the derivatives of the other five trig functions (without
resorting to the definition of the derivative!), because they are all related by simple algebraic equations. For
example, sin2 x + cos2x = 1 so if we differentiate implicitly (and use ddx sin x = cos x) we get:
2 sin x (cos x) + 2 cos x ddx cos x = 0 hence we can solve for
ddx cos x = - sin x . Similarly, tan x =
sin xcos x so
ddx
tan x = cos x (cos x) - sin x (- cos x)
cos2x =
1
cos2x (QUOTIENT rule) hence
ddx tan x = sec
2x . Also,
sec x = 1
cos x so ddx sec x =
ddx (cos x)
-1 = - (cos x)-2 (- sin x) = sin x
cos2x which can also be written
ddx sec x = sec x tan x . Finally, we "leave it as an exercise" to verify that
ddx csx x = - csx x cot x and
ddx cot x = - csc
2x .
Note that all the co-functions (cosine, cosecant and cotangent) have the (-) sign ... and that makes it somewhateasier to remember.
the EXPONENTIAL and LOG functions
• To find ddx 2
x requires the limit:
limh->0
2x+h - 2x
h and ultimately the weird limit
limh->0
2h - 1h .
• To determine ddx log3 x requires, ultimately, evaluation of another weird limit, namely:
45
limt->0
(1+t) 1/t = e ≈ 2.71828 (... which is how "e", the base of the "natural" logarithms arises).
Let's start with f(x) = loga x (for some positive base "a").
Then f '(x) =
limh->0
loga(x+h) - loga x
h =
limh->0
1h loga
x+hx =
limh->0
1x loga
1+hx
x/h
where we've used two magic properties of logarithms: log A - log B = log AB and n log P = log P
n. In the above, we
must assume that x > 0 (since loga x isn't defined for x ≤ 0) and that means we won't then get into trouble with
the factor 1x . Now, for any positive x, as h->0, then t =
hx ->0 as well. Hence our weird limit is
limt->0
(1+t) 1/t ,
just as we said. The values of (1+t)1/t, as t->0, approach a number slightly less than 3 (approximately2.7182818) and this number is called "e".
limt->0
(1+t)1/t = e or, equivalently,
limn->∞
(1+1n)n = e
Finally, then, we have ddx loga x =
1x logae
It's now clear that the world's greatest choice for a base for our logarithms is precisely the number e, since we'd
get: ddx loge x =
1x logee and since logee = 1 we'd have
ddx loge x =
1x . In fact, since this is the natural choice
for a base, logarithms to the base e ≈ 2.7182818 are called natural logarithms and are written ln x (pronounced"lawn x" or "ell en x").
ddx ln x =
1x
What could be simpler! Imagine the frustration of the pre-ln mathematicians who could find functions whose
derivatives were xn, for every number n ... namely xn+1
n+1 ... with the single exception of n = -1.
Anyway, from now on we'll assume our logs are natural logs unless otherwise specified.
Examples: Find dydx for each of the following:
(a) y = x ln x (b) y = ln cos x (c) y = ln (sec x + tan x)Solutions:
(a)dydx = x
ddx ln x +
d
dx x ln x = x 1x + ln x = 1 + ln x (using the PRODUCT rule for differentiation).
(b) Let u = cos x so y = ln u and (using the Chain Rule) dydx =
dydu
dudx =
1u (- sin x) =
1cos x (- sin x) = - tan x.
(c) Let u = sec x+tan x so y =ln u and dydx =
dydu
dudx =
1u ( )sec x tan x+sec2x =
sec x tan x + sec2xsec x + tan x = sec x.
Okay, we now have the derivative of logax. To find the derivative of ax we could try to evaluate the limit
limh->0
ax+h - ax
h = ax
limh->0
ah - 1h ... but it wouldn't be easy! Instead we digress to consider INVERSE functions
46
which will make it easier to find ddx a
x once ddx loga x is known.
PS:S: Wait wait wait. You said that log3 x isn't defined if x is negative.
P: ... or zero.S: Okay, log3 x isn't defined for x negative or zero. Why not?
P: Remember the definition of the logarithm to the base 3? If y = log3 x, then x =3y. But 3y is never negative or zero, so x can't
be negative or zero.S: There must be an easier way ...
P: Let's go over logs one more time. Suppose we want to find a number p such that 5p = 47. You can see that 52 = 25 and
53 = 125 so there is some magical number between 2 and 3 such that 5p is 47. Whatever that number, it's called ...
S: The log of 5 to the base 47?
P: No! p is the log of 47 to the base 5, written log5 47. Remember the significance of the word "base": if 5p = 47 then "5" is
the base. NOW, if there were such things as the log of a negative number, say p = log5(-47), then 5p = -47. But that's
impossible, right? You can't raise 5 to any power and get a negative number, so there are no logs of negative numbers.S: Or logs of zero, right?P: Right! Now let's talk about Inverse Functions. Pick a number between 1 and 10. Now multiply it by 3. Now subtract 7.
Now add 7. Now divide by 3. You get the number you started with, right?S: Amazing! How'd you do that?P: You pick a number x, then form f(x) = 3x-7. That is, you apply the function f(x). Then you undo what f(x) did to x, by
adding 7 and dividing by 3. That is, you apply another function g(x) = x+73
. See? You first apply the function
f(x) = 3x - 7, then you apply the function g and to get g(f) = f+73 =
(3x-7) +73
= x and the latter function undoes what f did,
and you get your "x" back again. The second function, "g", is called the INVERSE of "f". For the function "f" you "multiply
by 3 then subtract 7": f(x) = 3x - 7. For the function "g" you "add 7 then divide by 3": g(x) = x+73
... and g(f(x)) = x.
S: Why did you insist I take a number from 1 to 10?P: I didn't think you could multiply big numbers by 3.
LECTURE 6
INVERSE FUNCTIONS:If g(x) is the INVERSE of f(x), then g(f(x)) = x. Of course, in order that g(f(x)) make sense, x must lie in thedomain of f (because we've got an f(x) in there!), and f(x) must lie in the domain of g (since we're evaluating "g"at the place "f(x)"). How to find g, the inverse of a given function f?
Examples: • To determine the inverse of y = f(x) = 3x - 7, do the following:
(1) Write x = 3y - 7 (2) Solve for y = x+73 = g(x), the inverse of f(x).
Note: g(f(x)) = f(x)+7
3 = x.
• To determine the inverse of y = f(x) = x3:
(1) Write x = y3 (2) Solve for y = x1/3= g(x), the inverse of f(x).
Note: g(f(x)) = f(x)1/3 = x.
• To determine the inverse of y = f(x) = x
x+1 :
(1) Write x = y
y+1 (2) Solve for y = x1-x = g(x), the inverse of f(x).
47
Note: g(f(x)) = f(x)1-f(x) = x.
• To determine the inverse of y = f(x)(1) Write x = f(y) (2) Solve for y = g(x), the inverse of f(x).
Note: The first step is to interchange x and y: from y= f(x), you write x = f(y). This changes the graph ofy = f(x) by reflecting it in the line y = x.("Interchanging" replaces every point (x,y) by (y,x)which is the mirror image in the line y = x.) Thesecond step, solving x = f(y) for y in terms of x,doesn't change anything in the graph! In fact, x =f(y) and y = g(x) are two forms of the same relationbetween x and y ... and they have the same graph.This is handy. If you don't like the looks of afunction g(x), as in y = g(x), then you can write therelation as x = f(y) where the inverse function, "f",may be nicer to look at (and, in particular, todifferentiate!).
The last step, "solve for y", may be impossible ...because not all functions have inverses.
Consider y = x2. To find the inverse we interchange x and y,
writing x = y2. Then we solve for y = ?!%#*, but there are TWO
values of y, namely y = ± x , so x = y2 doesn't define y as a
function of x. That's because f(x) = x2 doesn't have an inversefunction. We might have anticipated this by considering the
graph of y = x2. When we interchange x and y, hence reflect thecurve in the line y = x, we don't get the graph of a function. (Afunction must have ONE y-value for each x-value in its domain,
and x = y2 doesn't!)
Remember the vertical line test for a function? Every vertical line (in itsdomain) must cross the graph only once. In order for a function to have aninverse, it must also satisfy a horizontal line test: every horizontal line (in
its range) must cross the graph only once. y = x2 satisfies the first test (soit's a function) but not the second (so it has no inverse). Remember: to findthe inverse of a function f(x) we must solve x = f(y) for y (or, what isequivalent,
solve u = f(v) for v ... or w = f(q) for q ... don't get tied to any particular labels!). To see if this is possible, we needn'tgo so far as interchanging x and y. We can simply ask: "Given y, is it possible to solve y = f(x) for x?" (i.e. canwe solve the original equation, y = f(x), for x, if y is given?) When y = 3 is given, it's like drawing a horizontalline y = 3 and trying to find the x that satisfies 3 = f(x). If there were one solution (and ONLY one) then there is
a unique x for y = 3 (and the horizontal line test is satisfied). (That's NOT the case with 3 = f(x) = x2.)However, if there were a unique x for all values of y in, say, c ≤ y ≤ d, then y = f(x) would define x as a functionof y (as well as defining y as a function of x!).What is this function? It's x = g(y) where "g" is the INVERSE of "f".
If f(x) has an inverse, g(x), then y = f(x), when solved for x, gives x = g(y)
48
TESTING TO SEE IF A FUNCTION HAS AN INVERSE:If the "horizontal line test" is satisfied for some function y = f(x) (on a domain a ≤ x ≤ b), then f(x) has aninverse. But the horizontal line test WILL be satisfied if the function f(x) is always increasing or alwaysdecreasing (i.e so-called "monotonic" functions)... and that can be used to test a function for an inverse: checkthat f '(x) > 0 or f '(x) < 0 on the domain.
More Examples of Inverses:
• What is g(x), the inverse of f(x) = 1
x+2 ? (on x > 0 where we can check that f '(x) > 0)
Write y = 1
x+2 , interchange x and y so x = 1
y+2 , then solve for y = g(x) = 1x - 2.
The graphs of y = 1
x+2 and y = 1x - 2 are reflections of each other in the line y = x.
• What is the inverse of f(x) = e2x ? (on -∞ < x < ∞ where f '(x) > 0)
Write y = e2x , then x = e2y, then solve for y = 12 ln x.
The graphs of y = e2x and y = 12 ln x are reflections of each other, in the line y = x.
PS:S: If "g" is the inverse of "f", then what's the inverse of "g"?P: We can see the answer graphically. Start with y = f(x), reflect it in the line y = x and get its inverse, y = g(x). Now start with
y = g(x), reflect in y = x and get its inverse ...S: y = f(x), right? But is that a proof?P: Here's a proof: since g(f(x)) = x (that's what makes "g" the inverse of "f"), then apply the operation "f" to each side and get
f(g(f(x))) = f(x), or, to put different labels on things, f(g(u)) = u (where we've replaced f(x) by u). Now stare atf(g(u)) = u and realize that this makes "f" the inverse of "g": whatever "g" does to u, "f" undoes it, recovering u again.
S: You were about to say something clever about the derivative of ax, remember? You said knowing ddx
loga x makes it easier
to find ddx
ax ... then you digressed into inverses.
the Derivative of an Exponential Function:
If we write y = ax and want to determine dydx we can differentiate directly (but we'd have to use the definition of
the derivative) or we can write this same relation as x = loga y and find dydx by implicit differentiation. Since
we've already found the derivative of the log function (by using the definition) the latter scheme is easier.
y = ax fi x = loga y fi ddx x =
ddx loga y fi 1 =
1y logae
dydx (using the Chain Rule, since
ddx logay =
d
dy logay dydx and we know
ddy logay). Now we solve for
dydx =
ylogae
. As is usual with implicit
differentiation, the answer has y's in it. But y = ax so we get, finally: dydx =
ax
logae .
To recap (about the exponential and log functions): ddx a
x = ax
logae and
ddx loga x =
1x logae .
These involve logs to the base "a" but can be changed to "natural" logs as follows:
If p = logae then ap = e so (taking ln of each side) ln ap = ln e or p ln a = 1, so p =
1ln a . In other words,
logae = 1
loge a . That's not something special about logs to the base e. In fact: logab =
1logba
. We then have:
ddx a
x = ax ln a andddx loga x =
1x ln a
In particular, if we choose as base the number "e" (noting that ln e = 1):
49
ddx e
x = ex andddx ln x =
1x
Note that f(x) = ex is a function whose derivative is the same function! (Can you think of any others? How
about f(x) = π ex or f(x) = -47 ex? Clearly f(x) = C ex has this nice property, for any constant C ... and these arethe only functions with the property that f '(x) = f(x).)
Above is a reasonably accurate (computer-plotted) graph of y = ex and y = ln x. They are inverses one of the
other and knowledge of one (say y = ex) will tell you everything you want to know about the other. For example,
it's clear that e0 = 1, so ln 1 = 0 . Also, e1 = e so ln e = 1 .
Also, ln e-3 = -3 ln e = -3. Further, since limx->-∞ e
x = 0 , then lim
x->0+ ln x = -∞ .
Note that (as mentioned on earlier occasions) the logarithmic function is defined only for positive x.
LOGARITHMIC DIFFERENTIATION:Once in a while one has to differentiate exponential functions with variable exponents and variable bases. If the
base is a constant, say f(x) = 2x2, it's easy:
Let u = x2 then f(x) = 2u and dfdx =
dfdu
dudx =
d
du 2u
dudx = (2
u ln 2) (2x) = 2x 2x2ln 2.
If the exponent is constant it's also easy. For f(x) = (sin x5)3, dfdx = 3 (sin
x5)2 cos(x5) (5x4) = 15 x4 sin2x5 cosx5.
If both base and exponent are variables, it's NOT so easy ... unless we use "logarithmic differentiation":
LOGARITHMIC DIFFERENTIATION: Take the ln of each side ... then differentiate
Because ln f(x)g(x) = g(x) ln f(x) the result of taking logs is to eliminate variable exponents!
Example: Determine dydx if y = x
x.
Solution: First takes logs of each side ("natural" logs, of course) : ln y = ln xx = x ln x. Now find dydx using
"implicit differentiation": ddx ln y =
ddx x ln x gives
1y dydx = x (
1x ) + ln x = 1 + ln x. Now solve for
50
dydx = y (1 + ln x) = x
x (1 + ln x). (Note that the answer is definitely NOT given by ddx x
x = x (xx-1) because the
rule ddx x
p = p xp-1 only works when the exponent is a constant!)
Note: Taking logs before differentiation is a good scheme in many instances.
Example: Determine dydx if y =
(x+1)10x3
(x-1)2 .
Solution: Here logarithmic differentiation is also useful. Take the ln of each side, then find dydx implicitly.
ln y = ln (x+1)10x3
(x-1)2 = 10 ln (x+1) + 3 ln x - 2 ln (x-1) so
1y dydx =
10x+1 +
3x -
2x-1 hence
dydx =
(x+1)10x3
(x-1)2
10
x+1 + 3x -
2x-1
Example: Use the differential to estimate the percentage change in volume of a cubical box if there is a 1%error in the side length.
Solution: If V = x3, where x is the side length, then we write DVDx ≈
dVdx = 3x
2 so DV ≈ 3x2 Dx. This gives
an estimate of the change in volume when the side changes from x to x+Dx. To get a percetage change, we
divide by V = x3 giving DVV ≈
3x2 Dx
x3 = 3
Dxx hence the % change in volume is three times the % change in
side length.
S: Huh?P: That's what we mean by percentage change, isn't it? We divide the change in volume by the original volume ... and we can
multiply by 100 to get the % change. But let me show you something slick, using logarithmic differentiation (in case youwere wondering why this example is stuck in with logarithmic differentiation!).
Since V = x3 then ln V = ln x3 = 3 ln x and now we differentiate and get 1V dVdx = 3
1x and we use
DVDx ≈
dVdx and
get 1V DVDx = 3
1x so that
DVV ≈ 3
Dxx . The fractional change in V is 3 times that in x (and we could multiply
by 100 to get % changes). In fact, if we're interested in estimates of "fractional changes" (or percentagechanges) in some function y = f(x) when x changes by Dx, we should :
(1) take logs................................................ ln y = ln f(x)
(2) differentiate............................................1y dydx =
f '(x)f(x)
(3) replace dydx by the approximation
DyDx ...........
1y
DyDx ≈
f '(x)f(x)
(4) solve for the fractional change Dyy .............
Dyy ≈
f '(x)f(x) Dx
What we'd find is that the % change in the volume of a cubical box (measured in metres3 ) is 3 times the %
change in side length, and the % change in the area of a square (measured in metres2 ) is 2 times the % changein side length. Try it!
About Exponential Growth:One often hears the expression "Holy cow! It grows exponentially!" (or some such phrase). Indeed, the growth
of y = ex (or y = 2x or y = 10x) is explosive, although it wouldn't seem so from the above diagram. However, if
we plot y = ex for -100 ≤ x ≤ 100 the graph looks very much like the negative x-axis together with the positivey-axis! For negative x it vanishes almost immediately (to zero). For positive x it explodes (to infinity). Among
51
other things, this explains the "miracle of compound interest". If you stick $1000 in a bank which pays 12%
interest each year, then you'll have 1000 (1.12) or $1120 after a year and 1000 (1.12)(1.12) = 1000 (1.12)2 =
$1254.40 after 2 years and, after n years, 1000 (1.12)n dollars. The amount of money is an exponential function
of "n": M = 1000 an where a = 1.12 and, for a lifetime (where n = 75 years, say), the $1000 will grow to about$5 million dollars! (moral: start saving early and retire a millionaire.)
PS:P: That reminds me of the story of how the inventor of the game of chess was rewarded (or was it checkers?). The king asked
what she wanted and she said a grain of wheat on the first square, 2 on the second square, 4 on the next and so on, thenumber doubling with each square. The king laughed at the trivial request, but only for a moment. He pulled out his math
notes and quickly calculated that, on the last square alone, there would be 263 grains of wheat and that was more than all the
wheat in the kingdom. (263 is roughly 10,000,000,000,000,000,000. How do I know? Because 210 = 1024, or roughly 103,
and I can write 263 = (210)6.3 which is then roughly (103)6.3 which is roughly 1019.)S: If the exponential function looks like the negative x-axis together with the positive
y-axis, that means the log function must look like the ... uh, let's see ... thenegative y-axis together with the positive x-axis. Right?
P: Right.S: That means the log function grows very slowly.P: You got it.
About the number e:
The base of natural logarithms, the number e, is one of the most important numbers in mathematics ... perhaps
the most important after the number π*. It occurs in a variety of situations:• Radioactive substances "decay". (For example, radium eventually "decays", turning into lead). The amount of
the substance, S(t), after a time "t", satisfies an equation like S(t) = A e-bt where A and b are constants.
• The current in an electric circuit is often described by I(t) = A e-bt.
• Populations sometimes grow according to an equation N(t) = A ebt.
• The so-called "normal probability distribution" is described by e-b(x-c)2 (the infamous "bell curve").
• The amount of a substance entering into a chemical reaction is often described by AB( ebt - 1)
Aebt - A .
• If $A is left in the bank to accumulate interest at the rate of i% per year, the amount of money after n years is
A ( 1 + i
100 )n. If the interest is calculated monthly at a rate
i12 %, the amount after n years is
A ( 1 + i
1200 )12n. If calculated daily at a rate
i365 %, the amount would be A ( 1 +
i36500 )
365n (if there are
365 days per year). If the interest is calculated T times each year at a rate iT % (T=12 means "monthly" and
T=365 means "yearly") then the amount after n years is A ( 1 + i
100T )Tn . The big question: how much money
would you have after n years if the interest is calculated continuously (meaning T->∞). The answer is
limT->∞
A ( 1 + i
100T )Tn. To evaluate this limit we set N =
100Ti , or T =
iN100 and note that N->∞ as T->∞, so
our limit can be written:
lim
N->∞ A ( 1 +
1N )
inN/100 or A
lim
N->∞
( 1 +
1N )
N in/100
* Perhaps the third most important number is g, Euler's constant. It's defined by
g =
limnÆ∞
1 +
12 +
13 +
14 + ... +
1n - ln n ≈ .5772156649015
52
and we recognize the limit
lim
N->∞ (1 +
1N )
N = e. Hence the amount after n years is A ein/100 dollars.
For example, $1000 left in the bank for n = 1 year at 10% per annum compounded continuously will grow to
$1000 e.1 = $1105.17 compared to $1100.00 if it were compounded annually.
Examples:
Plot the graph of y = x2 e-x, showing where it's increasing or decreasing.
Solution:
Note first that y = 0 only at x = 0 (and is positive for allother x-values). Further,
dydx = x
2 (-e-x) + 2x e-x = x (2 - x) e-x
which is negative for x < 0 and again for x > 2 (hencethe function is decreasing there) and is positive for0 < x < 2 (so the function is increasing there).
Note, too, that
lim
x->-∞ x2 e-x = ∞ (since both x2 and e-x become infinite). Finally, to complete the picture, we should
compute
limx->∞
x2 e-x. Unfortunately, this is a problem where one factor becomes infinite and the other
becomes zero. Nevertheless, if we write y = x2
ex and have faith in the explosive growth of ex, it's not hard to
accept the fact that
limx->∞
x2 e-x = 0. We'll show this later when we consider limits which have the form 00 or
∞∞ (as is the case with
x2
ex ) . In fact, ex grows so rapidly (with increasing x) that
limx->∞
x1000
ex = 0 ... and you
could substitute any power of x and get the same "0" limit. (No matter how hard x1000 tries to get to infinity, ex
drags the fraction to zero.) On the other hand, the function ln x grows so slowly that
limx->∞
xp
ln x = ∞ for any
positive power p, no matter how small.
On the other hand, 2x is also an exponential function as is 10x and πx ... and they all grow very rapidly.PS:S: I thought you said that "e" was important. Is it more important than 2 or 10?
P: Well, if you want to know the truth, "e" usually occurs in the form of a function eax, for some constant "a". However, eax
can always be written 2bx where "b" is some constant ... so, in a sense, that makes "2" just as important as "e", doesn't it?
S: How's that? Is eax = 2bx? Is that what you're saying?
P: Sure. Let's do an example. I'll find "b" so that e3x = 2bx. First I take the ln of each side and get 3x = ln 2bx = bx ln 2 (using
that important log-property: log AB = B log A) and that means that b = 3ln 2
. See?
S: I really don't see why all the fuss about "e". Let's do something else, can we?
Example: The cost of manufacturing an item is $100 even if no items are manufactured ... and the costdecreases with each item; for x items the cost per item is 100 - .1x (i.e. the cost decreases by $0.1 for eachitem). Graph C(x), the cost of producing x items.
53
Solution: For x items the cost is C(x) = x (100 - .1 x) = 100 x - .1 x2
which is positive for 0 < x < 1000. (For x > 1000 the cost is negative andmakes no sense ... so we only consider x < 1000.) Further, C'(x) = 100 - .2x is first positive, for x<500 (the total cost increases) then negative, forx>500 (C(x) is decreasing). Finally, C''(x) = -.2 is negative so that C'(x) isdecreasing (meaning that the graph is concave downward).
It's clear that the maximum cost occurs for 500 items, and thistechnique of determining the maximum (or minimum) value of a functionby investigating its derivative is one we'll use later on.
S: Are you saying the cost of producing 1000 items is zero?P: That's what's called a mathematical idealization. Mathematicians do it all the time. When a mathematician shows you the
solution to a problem you can be sure that she's made some assumptions and you have to be clever enough to argue with theassumptions. Assuming the cost per item decreases by $0.1 per item is, of course, ridiculous. So pay attention and arguewith my assumptions! BUT, having made the assumptions, the math just carries on to the bitter end and knows little ofnonsensical conclusions.
Example: An orchard contains 240 apple trees, each tree producing 30 bushels of apples. For each additionaltree planted, the yield per tree decreases by 1/12 bushel (due to overcrowding). Sketch N(x), the total appleproduction as a function of x, the number of additional trees planted.
Solution: The total production is (number of trees)x(bushels per tree).For x additional trees, the number of trees is 240 + x and the bushels pertree is 30 - x/12. The total production is:
N(x) = (240 + x)(30 - x12 ) = 7200 + 10 x -
x2
12 and N'(x) = 10 - x6 is first
positive (until x = 60) meaning that the total production is increasing, thenN'(x) < 0 (for x > 60) so the production decreases. Note, too, that
N''(x) = - 16 < 0 so the curve is concave down. (It's clear that x = 60 more trees
should be planted to yield the maximum number of apples from theorchard.)
Example: Repeat the above if the number of bushels per tree is 10.All other values are the same.
Solution: The total production is
N(x) = (240 + x)(10 - x12 ) = 7200 - 10 x -
x2
12 and
N'(x) = -10 - x6 is decreasing for every choice of x ≥ 0. Again, N''(x) < 0 so
the curve is concave down. (Conclusion? don't add any more trees!)
ODDS 'n' ENDS ON CURVE SKETCHING:Even and Odd Functions:
For simple functions like f(x) = x2 or f(x) = x3 it is sufficient to sketch a function which passes through theorigin and is increasing with increasing x. However, certain properties make sketching a little easier.
Some functions, like f(x) = x2, have the same value for -x as for +x; that is, f(-x) = f(x). Such functions are
called EVEN functions. Some examples: cos x, 1+x2 - x4, x sin x and e|x|.
(I) If f(-x) = f(x) then f is an EVEN function
Other functions, like f(x) = x3, assume the opposite sign when x is replaced by -x. These are the ODD functions.
Some examples: sin x, x - x3 + x5, x cos x and sin x3.
(II) If f(-x) = - f(x) then f is an ODD function
54
When graphing these functions, it's only necessary to graph the function for x ≥ 0 since the graph for x < 0 canbe obtained via the above EVEN or ODD symmetry.
If neither of (I) or (II) holds, then the function is neither even nor odd. Some examples: 1 + x, sin (x+1), ex and,of course ln x (which only has values for x > 0).
Quick&Dirty Curve Sketching:If a picture is worth a thousand words then we should spend some time in sketching the graph of functions: y =f(x). We first find the places where f(x) changes from positive to negative (by crossing the x-axis, or perhapsjumping discontinuously across the x-axis), then the places where f '(x) is positive or negative (to see where thegraph is increasing or decreasing) then, if we're not exhausted, we find where f ''(x) is positive or negative (to seewhere the graph is concave up or concave down), then take note of any horizontal asymptotes (where x->∞) orvertical asymptotes (where y->∞). It's a lot of work!
Example: Sketch the graph of y = x
1+x2 .
Solution: Note that y = 0 when x = 0, and y < 0 when x < 0, and y > 0 when x > 0; the graph lies in the first
and third quadrants. In fact, f(-x) = -x
1+(-x)2 = -
x
1+x2 = - f(x) so the function if ODD.
Now consider dydx =
(1+x2)(1) - x(2x)
(1+x2)2 =
1-x2
(1+x2)2 . It's positive when x2 < 1 (i.e. in the interval -1 < x < 1) and
negative outside this interval.
We can also calculate d2y
dx2 =
-2x(3-x2)
(1+x2)3 and find that y' '>0 (hence is concave upward) when x lies in
- 3 < x < 0 and again in x > 3 . Elsewhere it's concave downward.There are no vertical asymptotes (since f(x) is neverinfinite for any value of x) but (dividing numeratorand denominator by the highest power of x)
limx->∞
1x
1
x2+1
= 01 = 0 so there is a horizontal
asymptote, namely y = 0 (the x-axis).
NOW ... if we're not interested in the details but
want an quick&dirty picture of y = x
1+x2 , we can
sketch it for very small x and for very large x.For | x | << 1 (meaning x is very small), we can
neglect the x2 in the denominator (compared to the
"1") so, approximately, y ≈ x1 = x, and we sketch
y = x (for x very small).
Next, for large x, we neglect the "1" compared with the x2 and get the approximation y ≈ x
x2 =
1x which we
sketch for very large x (i.e. |x| >> 1). Then we just join these pieces (praying that nothing too wild happens in
55
between small and large x).
Example: Sketch y = 1 - x + x2.
For small x, y is very nearly the straight line y = 1 - x (neglecting the x2).In fact, this is the tangent line at the origin!
Next, for very large x, y is nearly the parabola y = x2.
Example: Sketch y = x + 1x .
For small x we neglect the x term and get:
y = 1x (approximately) .
For large x we neglect the 1x term and get:
y = x (approximately).
These two curves, y = 1x and y = x, are easy to
sketch (for small and large x respectively) ... thenwe join them (and pray).
S: In the last two examples, how do you know that y = 1 - x + x2 and y = x + 1x don't cross the x-axis?
P: I said this was quick&dirty ... especially dirty. But, of course, we could check if 1 - x + x2 = 0 for any x-value, or if x +1/x = 0. In both cases, there are no roots of these equations ... but if we had to do too much work the method wouldn't bequick.
S: Then the graph of N = 7200 - 10 x - x2
12 begins (for x near zero) just like the straight line y = 7200 - 10x then decreases like
y = - x2
12 (for very large x) ?
P: Right.
S: And C = 7200 + 10 x - x2
12 starts out like the line y = 7200 + 10 x?
P: Mmm
S: And y = 1 - x3
x + x2 looks like y =
1-0x+0
= 1x for small x and like ... uh, y =
0-x3
0+x2 = -x for large x, right?
P: Mmm
S: And what about y = 1+sin x
x ? It looks like y =
1x for small x, and y =
1+ex
x ≈
1x when x is large and negative, and y
= (x + 1x ) sin
1x looks like y = x sin
1x when x is large and that looks like
sin 1x
1x
which looks like y = 1. And what about y
= ...P: zzzzz
56
LECTURE 7
MORE ON INVERSE FUNCTIONS
Recall that a function y = f(x) has an inverse on thedomain a ≤ x ≤ b provided it's monotonic there (i.e.f(x) is always increasing or always decreasing on a≤ x ≤ b). Further, the inverse (we'll call it g(x)),when graphed, is just the reflection of y = f(x) inthe line y = x. Note, too, that if the range of f(x) is c≤ y ≤ d, then this becomes the domain of the g(x),and the domain of f(x), namely a ≤ x ≤ b, becomesthe range of g(x). (All of this can be seen from atypical graph of y = f(x) and its reflection/inverse, y= g(x).)
Example: Verify that f(x) = x
1+x has an inverse on the domain 0 ≤ x < ∞, then find the inverse g(x) and its
domain and range.
Solution: First we compute f '(x) = (x+1)(1) - x(1)
(1+x)2 =
-1
(1+x)2 which
exists and is negative for all x in 0 ≤ x < ∞, hence f(x) has an inversethere (because it's monotonically decreasing for all x in its domain). Note,
too, that 0 ≤ x
1+x < 1 for 0 ≤ x < ∞; the range of y = f(x) is then
0 ≤ y < 1. We then have the domain of g(x) as 0 ≤ x < 1 and its range as0 ≤ x < ∞ (and we have these even before we find g!). To find g we first
write y = f(x) = x
1+x then interchange x and y, writing x = y
1+y . Now we
solve for y: cross-multiply
to get x + xy = y so x = y - xy = (1 - x) y hence y = x1-x which is the inverse: g(x) =
x1-x .
PS:S: Hold on! Are you sure that y = g(x) is the reflection of y
= f(x)? Shouldn't you check?P: No need to (unless I've made a mistake ... and I never
make a mistake). Nevertheless, here's the graph of both.See? A reflection ===>>>
S: But why did you restrict the domain of f(x) to0 ≤ x < ∞?
P: Because f(x) would be an increasing function on thisdomain and I could guarantee it had an inverse (andthat's why I didn't need to check the graphs of each to seethis ... I knew it would be okay).
S: Are you saying that no other domain would do?P: No, just that the one I picked would do.S: I'd like to see you pick another domain.
P: Okay, let's sketch the graph of y = f(x) (a picture is worth a thousand words, right?)
57
I note that, for x very small, y = x
1+x is much like y
= x (neglecting the x in the denominator). Also, I can seethat y = 0 when x = 0 and there's a vertical asymptote at x
= -1 and a horizontal asymptote y = 1 since
limx->∞
x
1+x =
1 (which I could also expect because y is much like x
0+x
= 1 when x is large ... neglecting the 1 in thedenominator). Now look at the graph. See? It's increasingfor 0 ≤ x < ∞?
S: Are you kidding? It's increasing everywhere!P: Well, that's true ...S: So you could have picked any domain for f(x) and you'd
still have an inverse. Go ahead, try the horizontal line test
on y = x
1+x . It never intersects more than once!
P: Remember we're looking for the inverse of a function f(x) on some domain. On any domain which includes x = -1,
f(x) = x
1+x isn't a function ... it doesn't have a single, unique value at x = -1 ... it doesn't have any value at all! We have to
start with a function, one which satisfies the vertical line test: every vertical line in the domain must intersect the graphprecisely once. If I choose a domain which includes x = -1 it doesn't intersect at all at x = -1. See?
S: Sounds like cheating to me. What does g(x) look like ifyou use the whole graph y = f(x)?
P: Let's reflect. It looks like this ===== >>>>>>>S: That's a perfectly good function, isn't it? Try your line
tests on it! Is the math so dumb it won't let me find aninverse for the whole f(x)? I could understand if therewere two y-values for each x, in g(x) I mean ... but justlook at the graph, I mean ...
P: Okay, okay. We could have looked for an inverse for the
function f(x) = x
1+x on the entire real line
- ∞ < x < ∞ with the exception of x = -1. Happy?S: I don't know why you made such a fuss ...P: Watch what can happen if we're not VERY careful. Let's
consider another function which is defined everywhereexcept at certain isolated points (like f(x) above wherethe isolated point is x = -1). And we'll also pick afunction which increases everywhere .
S: Except at those isolated points, right? I can hardly wait.P: I'll pick y = tan x. Like it?S: I don't even remember what it looks like.P: Like this ====>>>
And it does increase everywhere, except at those pointswhere it has no value (or, if you like, a vertical
asymptote), like x = - π2 and x =
π2 and so on. So do you
think it has an inverse on the entire real line:- ∞ < x < ∞ ? (except at the points of discontinuity, ofcourse).
S: No, because it doesn't satisfy the horizontal line test.P: That's good! In fact it doesn't have an inverse, unless we
restrict the domain to ...
S: Let me do it ... uh, to - π2 < x <
π2 , right?
P: Right! And the range is -∞ < y < ∞, so the domain andrange of the inverse will be ...
58
S: The domain is - ∞ < x < ∞ and range is - π2 < y <
π2 ... I
can see that. And I can also tell you what g(x) looks like... I'm using g(x) to represent the inverse of f(x) =tan x ... uh, is there a special name for this inverse?
P: Yes, it's called "arctan x".S: Okay, y = arctan x looks like this ====>>>
Now, let's see you find dydx
if y = arctan x.
the DERIVATIVE of ARCTAN X, the INVERSE TANGENT:
If y = arctan x and we wish to determine dydx we can rewrite this relation as x = tan y. (Remember, if f(x) and
g(x) are inverses, then y = f(x) is the same relation as x = g(y). If you "solve y = f(x) for x" you get the inverse
x = g(y) and, if you "solve x = g(y) for y" you'll again get the inverse y = f(x).) Now, to find dydx it's easier to find it
implicitly from x = tan y by taking the derivative of both sides: ddx x =
ddx tan y or 1 = sec
2y dydx hence
dydx =
1
sec2y and, as usual when using implicit differentiation, we get the derivative with y's in it. To obtain
dydx in
terms of x alone we need to find sec2y in terms of x, knowing that tan y = x. But sec2y = 1 + tan2y and tan y = xhence
sec2y = 1 + x2. Finally, then:
ddx arctan x =
1
1 + x2
Note that the derivative is always positive (indicating that the function arctan x is increasing, as it is!) and the
derivative = the slope of a tangent line, approaches 0 as x->∞ (i.e.
limx->∞
1
1+x2 = 0) as it should (judging from the
graph of y = arctan x). Further, we can use the Chain Rule to show that:
ddx
1
a arctan xa =
1
a2 + x2
the DERIVATIVE of ARCSIN X, the INVERSE SINE:If we attempt to define the inverse of sin x we run into the same difficulty as with y = tan x: the horizontal linetest isn't satisfied. However we can restrict the domain as we did with the tangent function so that this test issatisfied. The most natural selection is shown below.
Hence we consider y = sin x with domain - π2 ≤ x ≤
π2 and range -1 ≤ y ≤ 1. The inverse sine, denoted
59
y = arcsin x, will then have domain -1 ≤ x ≤ 1 and range - π2 ≤ y≤
π2 . Again, to determine
dydx from y = arcsin
x, we first rewrite this relation as x = sin y and differentiate implicitly: ddx x =
ddx sin y or 1 = cos y
dydx so
thatdydx =
1cos y . Once again we should eliminate y by finding cos y in terms of x, knowing that sin y = x. We have
cos2y = 1 - sin2y = 1 - x2 so that cos y = ± 1 - x2 and we have an ambiguity in sign!#$%*? If we choose
cos y = - 1 - x2 then dydx = -
1
1-x2 would be negative. However, it's clear that y = arcsin x has a positive slope, so
we choose instead cos y = 1 - x2 and get:
ddx arcsin x =
1
1 - x2
and, in general
ddx arcsin
xa =
1
a2 - x2
PS:
S: Hold on. Isn't the mathematics smart enough to pick out the sign for us? I mean, our inverse has a positive dydx
so why does
the math give us a choice?
P: We restricted the domain of the sine function to - π2 ≤ x ≤
π2 , remember? With this choice of domain our inverse will have
a positive dydx
. However ...
S: I got it! Had we chosen, say 0 ≤ x ≤ π, then our inverse would have positive and negative slopes.P: No! Choosing a domain 0 ≤ x ≤ π would give us a piece of the sine function that doesn't satisfy the horizontal line test ... so
it wouldn't even have an inverse.
S: Okay ... suppose we'd chosen a domain, say, π2 ≤ x ≤
3π2 . Then we would have an inverse and the derivative of this inverse
would have the negative dydx
... I see that now. One other thing; why the strange choice of names: "arctan" and "arcsin"?
P: I almost hate to mention this, but some people like to call them tan-1x and sin-1x because that's a common notation in
mathematics for "inverses". For example, the "inverse" of the number "2" is written 2-1 and the inverse of the matrix A is
written A-1 so it seems natural to call f-1(x) the inverse of the function f(x). I used to use the notation tan-1x and sin-
1x myself, then I started to use a computer algebra system (that actually does calculus) and to enter these functions at the
computer keyboard you have to type arctan x and arcsin x, not tan-1x or sin-1x (which isn't so easy on a keyboard).
Besides, there's enough in calculus that's confusing so it's better not to get students even more confused by using sin-1x
which some will want to rewrite as 1
sin x and that's quite wrong, of course.
S: Hold on ... did you say a computer that actually does calculus ? If computers can do all this stuff then why am I?P: You have a problem in economics or biology or kinesiology or physics and you turn it into a mathematical problem with
equations and functions which need to be differentiated, etc., and then you can give the computer the task of performing thedifferentiations, etc. ... but you have to do the first part.
S: You never explained why the funny name "arcsine".P: Remember the definition of sin A for any number A? You measure off an arc of length A, on the unit circle, and the y-
coordinate of the resulting point is sin A, that is, y = sin A. Now suppose you were given the y-coordinate (i.e. the value ofthe sine), and had to find the arc length A. Using our notation, A = arcsin y, so you'd ...
S: Don't tell me ... you'd be finding the "arc" of the "sine" or the arcsine for short. Cute. One last thing ... will we be findinginverses for the other four trig functions?
P: We could, if you're really interested.S: Will I have to know it for the final exam?P: No.S: Then I'm not really interested.
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Examples: Evaluate
(a) arcsin
-
12 (b) sin
arctan
- 32 (c) cos
arcsin
1
3
Solutions:
(a) Let A = arcsin
-
12 so sin A = -
12 . Then A lies in the range of the
arcsine function, namely - π2 ≤ A ≤
π2 . We choose an angle in this range
whose sine is - 12 ; clearly A is negative and is recognized as A = -
π3 .
(We use the convenient sin A = opposite
hypotenuse with opposite = -1 and hypotenuse
= 2, correctly placing the angle A in the fourth quadrant. In degrees, A = -60˚)
(b) Let A = arctan
- 32 so tan A = -
32 and A lies in the range of the
arctan function, namely - π2 < A <
π2 . Here we must choose the angle in
the fourth quadrant. It's not one of the familiar angles, but we only need to
know its sine, namely: sin A = sin
arctan
- 32 = -
3
7 = -
37 .
(c) Let A = arcsin 1
3 so sin A = 13 and A lies in the range of the arcsine
function, namely - π2 ≤ A ≤
π2 . We choose the angle in this range whose
sine is 13 ; clearly A is positive, lies in the first quadrant and has a cosine
of: cos A = cos
arcsin
1
3 = 83 .
Problem: Which of the following is the graph of:(a) y = sin (arcsin x)? (b) y = arcsin (sin x)?(c) y = tan (arctan x)? (d) y = arctan (tan x)?
61
Solution:
PS:S: Hold your horses! You said that if g(x) is the inverse of f(x), then g(f(x)) = x. So shouldn't we have sin(arcsin x) = x and
arcsin(sin x) = x and arctan(tan x) = x and ...P: But we defined arcsin x by first restricting the domain of sin x to - π/2 ≤ x ≤ π/2, remember? Else sin x wouldn't even have
an inverse. In the above problem I'm asking what the graph of arcsin(sin x) looks like without restricting x to lie in - π/2 ≤ x≤ π/2. For any number x (i.e -∞ < x < ∞) we certainly have a value for sin x and this value lies in -1 ≤ sin x ≤ 1 and this isprecisely the domain of the arcsine function so we can certainly compute arcsin(sin x) ... even without restricting x. If wedid restrict x to lie in the interval - π/2 ≤ x ≤ π/2 then we would have arcsin(sin x) = x. You just have to look at the graph ofarcsin(sin x) to see that.
S: What you mean is, if x is unrestricted then arcsin x is NOT the inverse of sin x so we shouldn't expect arcsin(sin x) to beequal to x.
P: Precisely!S: Then why didn't you say that? Anyway, how did you pick out the correct graphs?P: First, I know that sin(arcsin x) isn't even defined unless x lies in - 1 ≤ x ≤ 1, the domain of the arcsine function. That gives
me the first graph which only lies in -1 ≤ x ≤ 1. All the others are defined for all values of x in - ∞ < x ≤ ∞ ... except
62
arctan(tan x) where tan x doesn't even have a value at odd multiples of π/2. In fact, as x goes from just under π/2 to just overπ/2, tan x jumps discontinuously from -∞ to ∞ so arctan would jump discontinuously from arctan(-∞) = - π/2 to arctan ∞ =π/2 (remember the graph of y = arctan x?). That gives me the next graph. Next, as x goes from -∞ to ∞, sin x just oscillatescontinuously from -1 to +1 to -1 etc. so arcsin(sin x) would oscillate from arcsin(-1) to arcsin(1) to arcsin(-1) etc., that is,from - π/2 to π/2 to -π/2 etc. That gives me the third graph. Finally, as x goes from -∞ to ∞, arctan x goes continuously from- π/2 to π/2 (remember the graph of y = arctan x?) so tan(arctan x) goes continuously from - ∞ to ∞ (just like x did!).
S: But how did you know they would look exactly like the given graphs?P: I didn't (except for x in - π/2 ≤ x ≤ π/2, of course, where they would all be the same as y = x). But the problem is worded so
that these were exactly the graphs of four functions ... so I just had to identify which goes with which.S: Sounds like cheating to me.P: Not at all! If I said one of the following is the formula for the volume of a sphere of radius r, then you'd be able to pick
which one even if you didn't know the formula. After all, it's one of the given formulas.S: What formulas?
P: V = πr2, V = 4πr2, V = 4πr4 and V = 4π3 r3. Which is the volume of a sphere of radius r?
S: I haven't the foggiest ... wait, it's πr2. I remember that one, vaguely.
P: No, if r is measured in metres then πr2 is measured in metres2, so it's the area of something, not the volume of anything!See? Check the dimensions.
S: Okay, it's V = 4π3 r3 metre3, a volume. So what about V = 4π r3? It's measured in cubic metres. What's it the volume of?
P: Three spheres.S: Very funny.
Examples:
1. (a) Calculate dxdy if xy = arctan x.
(b) Prove that f(x) = ex - ln x has an inverse on x ≥ 1. If the inverse is called g, compute g'(e).(c) Calculate g '(-1), if g(x) = arcsin(x+1).
(d) Calculate h '(1), if h(x) = (x-1) tan (ex-1).
(e) Calculate arcsin(tan(arcsin( - 12 ))).
(f) Calculate dydx at (0,0), if arctan y = ln (sec x + tan x).
2. (a) Calculate dydx at x =
12 , if y = ln (arcsin x).
(b) Determine the equation of the tangent line to ln (x2 + y2) = arctan yx a t the point (1 ,0) .
Solutions:1.
(a)ddx xy =
ddx arctan x gives x
dydx + y =
1
1+x2 . Now solve for
dydx =
1
1+x2 - y
x .
(b) f '(x) = ex - 1x ≥ e - 1 when x ≥ 1, so f '(x) > 0 hence f has an inverse on x ≥ 1. From y = f(x) = ex - ln x we
interchange x and y and write x = f(y) = ey - ln y which, when solved for y gives y = g(x), the inverse of f(x).
Hence ddx x =
ddx (e
y - ln y) or 1 =
ey -
1ydydx so
dydx = g'(x) =
1
ey - 1y
. To determine g'(e) we need to know y
when x = e. But x = ey - ln y so, when x = e, we have e = ey - ln y which has the obvious solution y = 1 (since e1
- ln 1 = e - 0 = e). Substituting y = 1 into 1
ey - 1y
gives g'(e) = 1
e - 1 .
(c) g(x) = arcsin(x+1) gives g '(x) = 1
1 - (x+1)2 ddx (x+1) =
1
1 - (x+1)2 (where we've used the Chain Rule) .
Substitute x = -1 and get g '(-1) = 1.
63
(d) h(x) = (x-1) tan (ex-1) gives h '(x) = (x-1) ( )sec2(ex-1) ex-1 + tan(ex-1) and substituting x = 1 gives
h'( -1) = 0 + tan(e0) = tan(1).
It's just as easy to use the definition: h '( -1)=
l imx->1
f(x) - f(1)
x - 1 =
l imx->1
tan(ex-1) = tan(1).
(e) Let A = arcsin(- 12 ) so sin A = -
12 and A lies in the fourth quadrant. (In
fact, A = - π6 .) Then tan(arcsin(-
12 )) = tan A = -
1
3 (reading this from the diagram) and finally, the problem:
arcsin(tan(arcsin( - 12 ))) = arcsin(tan A) = arcsin
-
1
3 = ?.
As above, we can let B = arcsin
-
1
3 so that
sin B = - 1
3 and note that B also lies in the fourth quadrant.
Unfortunately B is not one of the garden-variety angles (0˚, 30˚, 45˚, 60˚, 90˚, etc. i.e. 0,π/6, π/4, π/6, π/2, etc.), so
we'd need a calculator to find the angle B. Punch in 1
3 and then ask for the arcsin and get the number
35.26438965 which is in the first quadrant, so we take its negative and get the answer - 35.26438965˚ which isin degrees (!?&%) so we either put our calculator in radian mode and do it all again ... or simply multiply by
π/180 to convert to radians. We get B = π
180 (- 35.26438965) = -.6154797084 (using π = 3.1415926535, or
some such approximation to this number).PS:S: Couldn't I just leave the answer as B = - 35.2644˚? (My calculator only has 6 digits.) Besides, I can never remember whether
I should multiply by 180/π or π/180.P: Sure ... just don't forget to indicate that it's in degrees. But remember: π/180 is π radians per 180 degrees and is therefore
measured in radians per degree so multiplying by degrees gives radians. Try it on 180˚. You'd getπ/180 (180˚) = π radians. See?
S: Then what'd I get if I multiplied, say 35˚, by 180/π?P: A very big angle ... and marks taken off.S: I never could understand why the big fuss about radians. Just why do you insist upon using radians instead of degrees?P: Remember the definition of sin A and cos A? We measure off a distance A along the circumference of a circle and the
coordinates of the terminal point are (cos A, sin A) ... and the central angle just happens to be precisely A provided itsmeasured in radian! But that's not all. When we come to differentiate f(x) = sin x we have to resort to the definition of thederivative and get
limh->0
sin (x+h) - sin x
h =
limh->0
2 cos (x+h2 ) sin
h2
h (using a magic trig identity: sin A - sin B = 2 cos
A+B2
sin A-B2
with A = x+h and B = x) so we
get: cos x
limh->0
sin h2
h2
, hence it reduces to the evaluation of a weird limit:
the limit of sin AA
as A->0, where A = h2 . Note that
sin AA
is an EVEN function of A (since
sin(-A)-A
= sin AA
) so we need only compute
lim
A->0+ ( )
sin AA
... the left-limit will be the same.
64
To evaluate this limit we consider the diagram whichdefines the angle "A" (for some small positive A) and usethe Squeeze Theorem. To do this we need to find
something both smaller and larger than sin AA
, each of
which has the same limit, then sin AA
will also have this
limit. The lengths of various lines and arcs are shown.The piece of the circumference has length A (as perdefinition). Also a side of the right-triangle within thecircle has length sin A (it is, after all, the y-coordinates ofthe point which defines the sine function). Finally weconstruct a tangent line at this point and extend it tointersect the x-axis.
This line segment has length tan A. (Remember tan A = opposite
adjacent and adjacent = 1 so opposite = tan A.) Now compare
the three lengths; we have sin A < A < tan A or sin A < A < sin Acos A
or since A is small and positive then sin A > 0 and we
can divide through by sin A (and not change the direction of the inequality!) to get: 1 < A
sin A <
1cos A
. Now take the limit
as A->0+ and get 1 ≤
lim
A->0+ ( )
Asin A
≤ 1 (since cos A->1 as A->0). That does it!
S: What about
limA->0
( )sin AA
? You got
limA->0
( )A
sin A = 1.
P: But
limA->0
( )sin AA
=
limA->0
1
sin AA
= 11 = 1.
S: And just where did you use the fact that A is in radians?P: If the central angle is A, in RADIANS, then the arclength is also A ... but ONLY if A is in radians. If you want to consider
the limit of sin AA
when the central angle is A degrees, then the limit will NOT be "1". In fact, let's suppose the central
angle is A degrees. Then what's the arclength?
S: It'd be ... uh, let's see, there's a formula: a = r q where "a" is the arclength and "q" is the central angle in radians and r is
the radius. Here r = 1 and our central angle, in radians, is q = π
180 A (where A is in degrees). So far so good?
P: Yes, good ... keep going ... and I'm glad to see you multiplied by π
180 and not
180π
.
S: Okay, the arclength would then be a = q = π
180 A, if A is in degrees. So what would that make the limit?
P: You're doing fine. Keep going.S: Let's see ... where are those three lengths ... yeah, I got it: sin A < A < tan A, except I'd have to put the correct arclength in
the middle, so it'd become sin A˚ < π
180 A˚ < tan A˚ (and I've even put in the "degrees" symbol, just to make you happy).
Now I'd want A˚
sin A˚ in the middle so I'd have to multiply through by
180π
1
sin A˚ and that'd
give me: 180π
< A˚
sin A˚ <
180π
1
cos A˚ and now I can let A go to 0˚ and put on the squeeze and get:
180π
≤
lim
A->0+ ( )
A˚sin A˚
≤ 180π
so the limit is 180π
... am I right? Wait ... the limit of sin AA
is π
180 , right?
P: Right! And if x is measured in degrees, what's ddx
sin x ?
S: Huh?P: We started all this by trying to differentiate sin x, remember? The derivative of sin x is
limh->0
sin (x+h) - sin x
h =
limh->0
2 cos (x+h2 ) sin
h2
h = cos x
limh->0
sin h2
h2
= cos x provided "x" is measured in RADIANS.
65
BUT, if "x" is in degrees, what's ddx
sin x ?
S: I'd say ... uh, ddx
sin x = π
180 cos x. Does that make sense?
P: Sure. We can interpret the derivative of a function as the slope of the tangent line to the graph of that function. So what
you've shown is that y = sin x, when plotted against x in DEGREES, has a slope at the place x which is π
180 cos x. Here's a
plot of y = sin x, with x in RADIANS:
It's not too surprising that the slope of the tangent line, at x = 0, is 1. Now here's a graph of y = sin x with x in DEGREES.What's the slope at x = 0?
S: Don't tell me! It's small ... it's 180π
... no, that's big ... it's π
180 , right?
P: Right. So do you still want to work in degrees? Remember, you'd have to use ddx
sin x = π
180 cos x, etc.etc.
S: Okay ... radians are great ... so when do we do something useful with all this stuff?
LECTURE 8
OPTIMIZATION PROBLEMS
ABSOLUTE MAXIMUM AND MINIMUM VALUES OF A FUNCTION:Having to find the value of x which maximizes some function, f(x), is a problem which occurs frequently. Thefunction f(x) could be a profit (which we'd want to maximize) or a temperature (so we'd be finding the hottestspot) or the strength of a beam (so we'd be finding the strongest beam) or the yield of apples from an orchard.Sometimes f(x) represents a loss (which we'd want to minimize) or expenses (which we'd also want to minimize)or perhaps the coldest spot in the lake (to see if the water will freeze). Calculus helps.Below we show several functions defined on some closed interval of the form a ≤ x ≤ b. Each function (exceptone) has both an absolute maximum and an absolute minimum value on [a,b] ... and we can see where theseoccur.
66
For h(x), the absolute min and max values occur at x = c and x = d respectively (where h '(x) = 0).For g(x), the absolute min occurs at x = c where g '(x) = 0 but the absolute max occurs at x = a.For f(x) the absolute maximum and minimum both occur at the endpoints x = a and x = b respectively.For F(x), the absolute minimum occurs at x = b but there is no absolute max.
To see this we should define precisely what we mean by "the absolute maximum value of a function on an interval".
The absolute maximum value of f(x) on an interval, is f(x0) where
(i) x0 is in the interval and
(ii) f(x0) ≥ f(x) for every x in the interval.
The absolute minimum value of f(x) on an interval, is f(x0) where
(i) x0 is in the interval and
(ii) f(x0) ≤ f(x) for every x in the interval.
Note that the maximum value must actually be a value of the function! For the function F(x) depicted in the
diagram above, the limit as x->a+ exists, but it isn't the absolute maximum value because it is NOT a valueachieved for any x in the interval. On the other hand, F(b) is the absolute minimum because F(b) ≤ F(x) for anyx in the interval and is actually achieved for x = b. Whereas all other functions depicted have both absolutemaximum and absolute minimum values, F(x) does not. Note that F(x) isn't continuous for every x in the intervalwhereas the other functions are continuous.Does that mean that continuity guarantees the presence of an absolute maximum and minimum? Look at thefollowing functions:
For y = 1 - | x - 1|, we have y = 1 - ( )-(x-1) = x when 0 ≤ x < 1 ... since | x - 1 | = - (x-1) ... and we also have y =
1 - (x - 1) = 2 - x when 1 ≤ x ≤ 2. This function is made up of two lines, it's continuous for all x in the interval[0,2] and has both an absolute maximum and an absolute minimum (which occur at x = 1 and x = 0 or 2respectively). The next function isn't continuous for all x in [0,2] and fails to have an absolute maximum,although it does have an absolute minimum (which occurs at x = 0, x = 1 and x = 2). It looks like continuity willguarantee an absolute maximum and minimum ... but look again:
The first function is f(x) = 1x on the interval 0 < x ≤ 1 and although it IS continuous for every x in this interval,
it still doesn't have an absolute maximum (although it does have a minimum at x = 1). Perhaps that's because ithas a vertical asymptote (i.e. becomes infinite)? No, because the second function, namely f(x) = x on the interval0 < x < 1 also has no abs. max (neither does it have an abs. min) ... and it's never infinite.So what are the criteria which guarantee that some f(x) will have both an absolute maximum and an absoluteminimum on some interval?
67
If f(x) is continuous on the closed interval a ≤ x ≤ b, then it has both anabsolute maximum and an absolute minimum on that interval.
The thing is, f(x) must be continuous and the interval must be closed (i.e. include both end-points). Look at allof the above graphs and convince yourself that this does indeed happen. This doesn't preclude a discontinuousfunction from having an absolute maximum and/or an absolute minimum ... even on an interval which is notclosed. It's just that we can't guarantee it. More than that, if a function doesn't have an abs. max or an abs. minwe can guarantee that the function is either discontinuous somewhere in the interval and/or the interval isn'tclosed.Suppose we do have a continuous function on a closed interval. Just where should we look to find the absoluteextrema? A study of the various graphs shown above (and below) leads us to consider places in the intervalwhere either f '(x) = 0 or where f '(x) doesn't exist.
Hence, we define a critical point:
If (i) f '(x0) = 0, or
(ii) f '(x0) does not exist,
then x0 is called a critical point of the function.
Then:
If f(x) is continuous on the closed interval a ≤ x ≤ b, then theabsolute maximum and absolute minimum will occur either(i) at a critical point in a < x < b, or(ii) at an end-point: x = a or x = b.
That means that we need only find the critical points interior to the interval and evaluate f(x) at each, thenevaluate f(x) at the end-points, then pick the largest and smallest of all these values. BUT, this procedure isguaranteed to succeed only if f(x) is continuous on an interval, and the interval is closed (as the followingexamples show).
Examples:
min & max min & max no max, but min occurs atat end-point and occur at each min occurs critical point,a critical point. end-point. at an end-point. but no max.
Examples:
For f(x) = 1 - | x - 1 |, the only critical point in 0 < x < 1 is x = 1 (because f '(1) doesn't exist). Hence the absolutemaximum and absolute minimum values are the largest and smallest of the values f(0), f(1) and f(2), hence theabs. max is f(1) = 1 and the abs. min is f(0) = f(2) = 0.For h(x), the abs. max and abs. min are the largest and smallest (respectively) of the numbers f(a), f(b), f(c) and
68
f(d) (noting that x = c and x = d are critical points in a < x < b because f '(x) = 0 there).For g(x), the abs. max and abs. min are the largest and smallest (respectively) of the numbers g(a), g(b), and g(c)(noting that x = c is a critical point in a < x < b because g '(x) = 0 there).
Example: Determine the absolute extrema of f(x) = x - 2 sin x on 0 ≤ x ≤ 2π.
Solution: f '(x) = 1 - 2 cos x = 0 where cos x = 12 . On 0 ≤ x ≤ 2π this occurs at x =
π3 and x =
5π3 . Hence
the absolute maximum and minimum values of f(x) are the max and min of the numbers: f(0), f( π3 ), f(
5π3 ) and
f(2π). Hence we calculate: f(0) = 0, f( π3 ) =
π3 - 2 sin
π3 =
π3 - 3 ª -.685, f(
5π3 ) =
5π3 - 2 sin
5π3
= 5π3 + 3 ª 6.968 and f(2π) = 2π ª 6.283, hence, the abs. min is
π3 - 3 and the abs. max is
5π3 + 3 .
Example: A conical drinking cup is formed from a circular piece ofpaper by removing a sector and joining the edges. If the radius of the piece of
paper is 10 cm., what should be the angle q so as to yield a cup ofmaximum volume?
Note: volume of a cone = 13 (AREA OF BASE) (HEIGHT)
Solution: From the diagram, the volume of the cone
is V = 13 π r
2 h. But 102 = r2+h2 (that's Pythagorus talking) so
r2 = 100 - h2 hence the volume is V(h) = 13 π (100-h
2)h and we wish
to find the absolute maximum of this continuous function for h insome interval ... hopefully a closed interval. Clearly the smallesth is 0, but that means we don't cut any sector out of the circularpaper, so q = 0, and we don't have a cone at all. On the otherhand if we cut out
all the paper, so q = 2π, then although h is a maximum (h = 10) we again don't have a cone. It seems we have an
open interval to consider, namely 0 < h < 10. Nevertheless, we continue: V'(h) = 13 π (100 - 3h
2) which is first
positive, for 0 < h < 10/ 3 (so V(h) is increasing) , then negative, for 10/ 3 < h < 10 (so V(h) then decreases)
and the maximum clearly occurs at the critical point, h = 10/ 3 , where V'(h) = 0. Since we were asked for q,we must find the relationship between h and the angle q.Note that the circumference of the circle (on the cone) is 2πr and this equals the remaining circumference on thepaper circle, namely: 2π(radius) - arc length = 2π(10)-10q = 20π - 10q ... using arc length=(radius)(angle at
centre). Hence, 2πr = 20π-10q so q = 2π - π5 r but r = 100-h2 = 100 -
1003 = 10
23 so q = 2π - 2π
23
.Finally, we really could have considered V(h) on the closed interval 0 ≤ h ≤ 10 (even though the "cone" wouldhave zero volume at the end-points: h = 0 and h = 10). However, we pretended that this was an open-intervalproblem so we could talk about ....
RELATIVE MAXIMA and MINIMA:The previous example illustrates how we might obtain max and min valuesof f(x) on intervals which are NOT closed: use the derivative to determinewhere f(x) is increasing or decreasing. Indeed, if f '(x) > 0 just to the leftof x = c (meaning that f(x) is increasing in value) and f '(x) < 0 just to theright of x = c (so f(x) is decreasing), then we might suspect that f(c) islarger than nearby values of f(x). That should make f(c) a RELATIVEMAXIMUM. Of course, there's always the possibility that f(x) isdiscontinuous at x = c and
the actual value, f(c), is different from the left- and right-hand limits. Then f(x) wouldn't have a derivative at x = c,
69
but that would still make x = c a critical point, so we again look to the critical points for RELATIVE MAXIMAand RELATIVE MINIMA.For the function shown at the right (which is the previous function, nowgraphed over all of (a,b]), f '(c) = 0 so x = c is a critical point (hence acandidate for a relative max or min) so we check the sign of f '(x) just tothe left and right of x = c and find that f(x) first increases then decreases.Although this makes f(c) a relative maximum, it's clearly not the absolutemaximum on the interval shown, since f(b) is. For the function depicted,there are two relative minima and three relative maxima in the interval.(Can you find them?)
For the previous function, there is no absolute minimum at x = a because f(x) doesn't have a value there(indicated by the open circle). For a similar reason, there isn't a relative minimum at x = a either.It's time we defined relative extrema:
f(c) is a relative maximum if f(c) ≥ f(x) for all x in the domain which are sufficiently near c
If you're standing on the top of a mountain you're at a relative maximum (of elevation). To be at an absolutemaximum, the mountain has to be Mount Everest.Now, we consider the following "First Derivative Test" for a relative maximum:
If f '(x) > 0 immediately to the left of x = c and f '(x) < 0 immediately to the right of x = c, then f(c) is a relativemaximum??Note the ?? Is this test valid? Can we think of a counterexample? (i.e. a function which is increasing just to theleft of x = c, decreasing just to the right, yet f(c) is NOT a relative maximum?)On the right is the graph of a function, f(x), almost identical to the oneportrayed previously, yet f(c) is no longer a relative maximum. Indeed,f(c) is now a relative (and absolute) minimum! Hence, if f(x) increasesuntil x = c then decreases after x = c, it's not enough to guarantee amaximum (of either flavour). We need continuity at x = c. (That is not tosay that f(c) cannot be a relative or absolute maximum if f(x) isdiscontinuous at x = c, it's just that we can't guarantee it!)So we modify our test:
First Derivative Test for a MaximumIf f(x) is continuous at x = c and f'(x) > 0 immediately to the left of x = c and f'(x) < 0
immediately to the right of x = c, then f(c) is a relative maximum.
First Derivative Test for a MinimumIf f(x) is continuous at x = c and f'(x) < 0 immediately to the left of x = c and f'(x) > 0
immediately to the right of x = c, then f(c) is a relative maximum.This theorem must be taken with a grain of salt. i.e. it's true, except it doesn't include some important cases. Forexample, if a relative maximum occurs at x = a (the left end-point of our domain), then it's pretty hard to insistthat f '(x) > 0 to the left of x = a. Similarly, for the previous graph, f(b) wouldn't satisfy the requirements of thistest even though it certainly is a relative maximum. Clearly, for end-points, we need continuity and theappropriate sign for f '(x) on one side of the end-point. It just makes sense to say: "If f(x) increases to the valuef(b) at the right end-point, then f(b) is a relative maximum for f(x) on the interval".Note, too, that if f(c) is a relative maximum, we needn't have
f '(c) = 0. It may be that f '(c) doesn't exist as is the case for f(x) = 1 - |x-1| at x= 1. However, if f(c) is a relative maximum, then either f '(c) = 0 or itdoesn't exist. In either case it's a critical point.We now have a scheme for identifying relative maxima and absolutemaxima. To reiterate:If it's required to find extrema (either relative or absolute) for f(x) on some interval, then we first determine thecritical points of f(x) interior to the interval. (We'll consider the end-points separately.) If f(x) is continuous at
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such an interior critical point then we apply the First Derivative Test to see if it's a relative maximum orminimum. We also evaluate f(x) at each critical point. Then we evaluate f(x) at the end-points. If f(x) iscontinuous throughout the closed interval [a,b], the absolute maximum and minimum are simply the largest andsmallest values of f(x) evaluated at the end-points and the interior critical points.
S: What if f(x) isn't continuous throughout ... or what if f(x) isn't even continuous at some critical point? Then what? Yourtheorems seem pretty weak to me.
P: Actually, these theorems are only for the very lazy who like to turn some crank and have the answers pop out. The best wayto determine the absolute and relative extrema is just as we've been doing with our graphs. We just look ... and see. After all,a picture is worth ..
S: Yeah, I know, a thousand words.P: Here's a nice problem where the graph tells all. Take a length of wire, of
length, say, 1 metre. Cut it into two pieces. With one piece, form acircle. With the other, form a square. How much of the wire should formthe circle and how much the square if you want the maximum total area?Okay, suppose we cut a length x metres for the circle and the rest, 1-x,
for the square. The area of the square is ( )1-x4
2 and the area of the
circle is πr2 where r is the radius. Since the circumference
is x, then x = 2πr so r = x2π
so the area of the circle is π( )x2π
2. The total area (which we wish to maximize) is:
A(x) = π( )x2π
2 + ( )
1-x4
2 and we now graph A(x) by determining where it's increasing and decreasing. We have A'(x) =
x2π
- 1-x8 =
(4+π)x - π8π
which is negative at first (when x is small) then positive (when x is larger than π
4+π ).
The graph then looks like this ====>>>
The absolute minimum obviously occurs at x = π
4+π and the absolute maximum
occurs at one of the end-points. To see which, we evaluate A(0) = 116
and A(1)
= 14π
. The latter is larger, so the maximum area is achieved when x = 1; ALL of
the wire is used to form the circle.
S: But A(x) is continuous on 0 ≤ x ≤ 1 (which, by the way, is a closed interval) so you can find the absolute maximum byevaluating A(x) at the end-points and at the interior critical point, right?
P: Right.S: Then you didn't need to graph A(x) at all ... or determine where it was increasing or decreasing. Just pick the biggest of
A(0), A(π
4+π ) and A(1). Am I right?
P: Yes ... and see how much you've learned already? I'll bet if you did this problem last month, you'd set A'(x) = 0, find that x
= π
4+π and conclude that that gave the maximum area ... just because I asked for the maximum. If I had asked for the
minimum area, you'd do the same thing. Or maybe you'd apply some kind of test, or ..S: Certainly not. And now, can we continue?P: Just one last thing. If you had used the second piece of wire for an equilateral triangle, or perhaps some other figure, do you
know what the solution would be? I'll tell you. It would STILL be x = 1; ALL of the wire should be used for the circle. Doyou know why? I'll tell you. The circle provides the maximum possible area for a given perimeter ... so you shouldn't wasteany wire on anything else. In fact, for a given surface area, what solid provides the maximum possible volume? I'll tell you:a sphere. That's why soap bubbles are spherical. In fact, every soap bubble has taken this math course and ...
S: Please, let's keep going.
Example: For each of the following, determine where the function is increasing or decreasing, locate critical points, determine the location of relative maxima and relative minima and use all of this information to sketch the graph of y = f(x).
(a) f(x) = x5 - 5x
(b) f(x) = x ex (note: to help in graphing, use lim
x->-∞ x ex = 0)
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Solutions:
(a) f '(x) = 5(x4 - 1) ≤ 0 (i.e. f(x) is decreasing) where x4 ≤ 1, i.e. where -1 ≤ x ≤ 1. Elsewhere,
f '(x) > 0 (and f(x) is increasing). Critical points occur where f '(x) = 5(x4 - 1) = 0, hence at x = ± 1. Since f '(x) isfirst positive then negative as x crosses -1, f(-1) is a relative maximum. Similarly, f '(x) is first negative thenpositive as x crosses +1, so f(1) is a relative minimum. (Also, for x small, f(x) behaves like y = - 5x.)
(b) f '(x) = (x+1) ex ≤ 0 when x ≤ -1 (so f(x) is decreasing there) and f '(x) > 0 elsewhere (so f(x) is increasing).The only critical point occurs at x = -1 (where f '(x) = 0). Also, f(x) < 0 when x < 0 and f(x) > 0 when x > 0, so
the graph lies in the first and third quadrants. Finally, since lim
x->-∞ x ex = 0, y = 0 is a horizontal asymptote.
S: How did you know that lim
x->-∞ x ex = 0?
P: I write it as lim
x->-∞ x
e-x and use the explosive growth of the exponential function (since the exponent approaches +∞) to
drag the fraction to zero as x->-∞.S: Is that a proof?P: No ... but later we'll discuss a method for evaluating such a limit.
In the "wire problem", we wanted to cut the wire in an "optimum" manner (to maximize the total area). Whenwe chose the variable "x", we did so knowing that when we found x it would determine precisely how to cut thewire. That's important. If we want to maximize a quantity Q, then choose a variable, call it x (or someconvenient name), which will determine precisely how to maximize Q ... then express Q in terms of x; for each xthere is a Q-value. Then find the x-value which maximizes (or minimizes) Q(x).For example, if the problem were to determine the isosceles triangle ofmaximum area which can be inscribed in a given circle of radius R, wewould NOT pick as variable the base length r. Why not? Because even ifwe knew the "optimum" value of r it still wouldn't tell us what triangle touse! In fact, there are TWO triangles for each value of r. (See diagram===>>>). Moral? Pick a variable (such as the height or the angle at thevertex) which DOES provide a unique triangle.
Example: A man can run 10 times faster than he can swim. Hebegins in the water at a point P (see diagram), swims to shore,then runs to Q (the cottage). Describe his path so the total time isa minimum.Note: distances "a", "b" and "c" are given.
Solution: Since the total time is to be minimized, we let it be
called T. Then we pick some variable which describes a (unique!)path from P to Q and express T in terms of this variable. Then,minimize T.There are several possible variables to describe a particular path fromP to Q: the distance x or the distance y or the angle q or the angle f.(If any of these is known, then the precise path will be known, hencethe time T). Let's pick x. Then the total time, T(x), is made up of thetime spent in the water + the time on land. We calculate each:
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We use time-in-water = distancespeed =
a2+x2
u where we let u = speed-in-water (since we weren't given this
quantity, we simply give it a name and use it!) Note that the distance travelled in the water is calculated fromPythagorus' theorem.
Also, time-on-land = distancespeed =
b2+y2
v where v = speed-on-land. Hence, the total time is:
T = a2+x2
u + b2+y2
v . As is often the case, our function depends upon two variables, x and y, which are related.
We must eliminate one of them so T is a function of a single variable. Since we chose to express T in terms of x,we'll eliminate y. There must be a relation between x and y ... and there is. It's x + y = c (which was given).
Hence, y = c - x so we have, finally, T(x) = a2+x2
u + b2+(c-x)2
v and we remind ourselves that a, b,c,
u and v are constants which we assume are known. Our problem is to minimize T(x) for x in some interval ... butwhat interval? It's clear that x lies in 0 ≤ x ≤ c (since the swimmer would head for shore, somewhere between Pand Q) hence we have the problem of a continuous function T(x) on a closed interval [0,c]. We find the criticalpoints:
T '(x) = 1u
x
a2+x2 -
1v
c-x
b2+(c-x)2 = 0 and we must now solve this for x. If it should turn out that x lies in (0,c),
then it's an interior critical point and we evaluate T(x) there, as well as at x = 0 and x = c (the end-points). Thesmallest of these three numbers is the minimum time. Note how convenient to have the theorem which says wedon't need any derivative test if T(x) is continuous and the interval is closed ... just pick the smallest of these
values! Without this we'd have to analyze the terrible expression T '(x) = 1u
x
a2+x2 -
1v
c-x
b2+(c-x)2 to see
where it's positive and where it's negative. Or, worse yet, we might be tempted to use the "second derivative
test" which requires finding d2T
dx2 !! It will turn out (of course!) that T '(x) = 0 at one point in the interval
0 < x < c and T '(x) < 0 to the left of this point and T '(x) > 0 to the right, so this point provides the minimum.
Let's get back to the equation which gives the critical point: 1u
x
a2+x2 =
1v
c-x
b2+(c-x)2 or
x
a2+x2
c-x
b2+(c-x)2
= uv which, although it looks messy, is actually a very nice way to express the condition for a minimum
time because each of x
a2+x2 and
c-x
b2+(c-x)2 is recognized (!after considerable staring at the
expressions!) as the sine of one of the angles q and f !! The condition for a minimum can then be written
sin qsin f =
uv and, for our problem, the ratio of speeds is
speed-in-waterspeed-on-land =
110 so we didn't have to know each speed
after all, just their ratio! This magnificent condition for the minimum time is called Snell's law and is known toall beams of light ... because it's precisely how light travels from one medium to another! When light movesfrom glass to air (where it travels faster) it is refracted at the boundary between the two media (just like the pathof our swimmer ... because the speed of light is greater in air than in glass). Having taken this math course, thelight beam changes course at the boundary so it takes the minimum time to travel from one point (in glass) toanother (in air).
PS:S: You haven't finished the problem! So what's the optimum path? So what's x?
P: Well, I need to solve
x
a2+x2
c-x
b2+(c-x)2
= uv =
110
.
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S: Go ahead. I'll wait.
P: Okay ... I'll let r = uv =
110
, the ratio of speeds (so I only have to do this once and when I'm finished you can stick in any
ratio you like). Then I square both sides and get an equation to solve for x, namely ... uh, I think it's a quartic equation: (1-
r2) x4 - 2c (1-r2) x3 +(b2+c2(1-r2)-a2) x2 + 2ca2r2 x - a2c2r2 = 0. See? Once you know all the numbers a, b, c and r ... youjust solve this equation for x.
S: You gotta be kiddin' ... just solve for x? Besides, how do I know you didn't make a mistake. Maybe your quartic equation iswrong.
P: Well, we can check it ... sort of. If x is measured in metres, then every term must have the same dimensions (else I did make
a mistake). Since r is the ratio of speeds, it has no dimensions (i.e. metres/secondmetres/second
is dimensionless.) Now look at each term.
(1-r2)x4 has dimensions m4, and since a, b and c are also in metres, then 2c (1-r2) x3 and (b2+c2(1-r2)-a2) x2 and 2ca2r2 x
and finally a2c2r2 are also in m4. So that checks out. Now let's assume r = 1 (so the speed is the same in water and on land)
and b = a. Then we know the solution: the path should be a straight line from P to Q and x = c2 must be the solution.
Our equation becomes: 2ca2x - a2c2 = 0 and x = c2 is indeed the solution! So that checks out. Hence I have great faith in
my equation ... I don't think I made a mistake.
S: Okay, solve it if a = 1, b = 2, c = 3 and r = 110
.
P: The equation becomes: 99 x4 - 594 x3 + 1290 x2 + 6x - 9 = 0 and I'll use Newton's method to find x. Aaah, wait, you don'tknow Newton's method, do you? You'll just have to wait for a lecture or three until we get to that topic.
S: Sure, sure. But tell me, what's this about checking the dimensions? You've mentioned this before.P: It's a neat idea. If somebody gives you a formula for something, you can check that it's dimensionally correct. If not, then the
formula is wrong. Suppose somebody says the area of a circle is πr3. Then, for r in metres, this would give cubic metres for
the area (because of the r3) so it can't be a correct formula since area is measured in square metres. Suppose somebody says
the volume of a torus (that's a donut) is V = πa2b2 where "a" and "b" are
certain lengths, then it's an incorrect formula because it gives m4, not m3 as
it should for a volume. Suppose Einstein had said E = mc3. Then you couldcheck and discover that the dimensions of energy E are not the same as the
dimensions of mc3, so Einstein was wrong. In fact you could point out to
him that he's better off with E = mc2 because that, at least, is dimensionally
correct. You see? A = πr2 is the area of something and C = 2πr is the length
of something and V = 4π3 r3 is the volume of something, etc. so you would
never say the area of a circle is 2πr because it's not even an area!
S: But what if I said the area of a circle is A = 2πr2? It has the correctdimensions, so then what?
P: Then checking the dimensions wouldn't help. But, sometimes, you can check particular cases too, as we did with ourequation above. We knew the answer when r = 1 and a = b. It's the same for, say, the volume of a torus. If I said the volume
was V = π2
4 (a+b) (b-a) 2 then it not only has the correct dimensions (metres3 if a and b are in metres) but it also gives V = 0
when a = b (so, for example, V = π2
4 (a+b) 3 would definitely be wrong).
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S: Is it the correct formula for a torus?P: Yes, I think so, but wait until we get to volumes later on in the course; we'll compute the volume from scratch.S: Newton's method comes later ... and volumes comes later ... and I thought we were almost finished. Okay, let's go!
Example: A page is to have 100 cm.2 of printed text, with 2 cm. borders.Find the dimensions of the page of smallest area.Solution: We must minimize the area of the page, so we give it a name: call it A.
Now we pick some variables which will determine the dimensions of thepage: say x = width and y = height. Then A = xy must be a minimum. Asbefore, we have two variables, but they're related since the width of thetext is (x-4) and its height is (y-4) and its area is given as (x-4) (y-4) =
100. Now we can find y = 4 + 100x-4 and substitute to get
A(x) = xy = 4x + 100xx-4 and that's
the function we wish to minimize. Unfortunately, we don't have a closed interval because x can be any length greater
than 4 cm. (If x = 4.0001, then we have little width for our text, but y = 4 + 100x-4 is enormous so the page is a
mile high!) Okay, we'll find A'(x) and see when it's increasing and when it's decreasing:
A'(x) = 4 - 400
(x-4)2 which is VERY negative when x is slightly greater than 4
(example: x = 4.0001) and positive when x is very large (since the second
term is small). Hence A(x) is first decreasing, until A'(x) = 4 - 400
(x-4)2 =
0, then increasing. The minimum occurs when (x-4)2 = 100 or x - 4 = 10,
hence x = 14 cm. (and y = 4 + 100x-4 = 4 + 10 = 14 cm. as well).
If we wanted to sketch A(x) (although it's not necessary, since we have the
answer) we could note that A(x) looks like 100xx-4 when x is near
4 (neglecting the first term, 4x), so A(x) has a vertical asymptote ... and A(x) looks like 4x when x is large(neglecting the second term).
LECTURE 9
RELATED RATE PROBLEMS
Many times we know the rate of change of some quantity and wish to know the rate of change of a relatedquantity; that's a "related rates" problem. For example, if the radius of a balloon is increasing at 10 m/second,
how rapidly is the volume increasing? We know drdt (r is the radius) and we wish to know
dVdt (V is the volume)
. Identifying these rates of change is the first step. The second step is to find the relation between V and r: V =4π3 r3. Then we differentiate both sides with respect to t and get
dVdt = 4π r
2 drdt , hence we can compute
dVdt for
any radius r, since drdt = 10 is known. A typical problem is worded: "If the radius changes at the rate 10 m/s then
find the rate of change of volume when the radius is 2 m." See? Both drdt and r are given and we just plug them
into dVdt = 4π r
2 drdt to get
dVdt .
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Example:Sawdust is falling onto a pile at the rate of K metre3/second. If the pilemaintains the shape of a right circular cone with its height equal to thediameter of the base, how fast is the height increasing when the pile is Hmetres high?
(Note: your answer will be in terms of numbers K and H.)
Solution: We identify what rate of change is requested: dhdt where h is the height (in meters). The rate of
change which is given is dVdt = K m
3/s. (V is the volume of the cone.) Next, we find the relation between V and
h.The volume of the conical pile (when the height is h and the radius is r)
is 13 π r
2 h (depending upon two variables). But r = h2 (given) , so the volume
is V(h) = π12 h
3 metres3 and is changing at the rate:
dVdt =
dVdh
dhdt =
π4 h
2 dhdt = K m
3/sec. (given). Hence dhdt =
4K
πh2 m/sec. and,
when h = H, we have dhdt =
4K
πH2 m/sec. (We can also check that the
dimensions of this quantity are: m3/s
m2 = m/s which is okay for dhdt .)
Example:
Water leaks from a conical tank at the rate A m3/minute. If the tank is Hmetres high and R metres in radius (across the top), how rapidly is thedepth of water changing when the tank is half full? (Express your answerin terms of A, H and R .)
Solution:
Again we note that dVdt = A m
3/min is given and dhdt is required and the
relation is again V = 13 π r
2h. But the relation between r and h is obtained
by similar triangles: rh =
RH so r =
RH h hence V =
13 π
R2
H2 h3 (which has the dimensions m3 ... which is comforting).
Now dVdt = π
R2
H2 h2 dhdt = A m
3/min. (given) so we need to find h when the container is half full,
substitute into π R2
H2 h2 dhdt = A and solve for
dhdt . We'll interpret "half-full" to mean
half the volume of the full container. But when h = H, the volume is 13 πR
2H so we
want to know h when V = 13 π
R2
H2 h3 =
1
2 13 πR
2H. Solving we find that h3 = H3
2
so h = H
21/3 (which has the correct dimensions!) Using this h we get
dhdt =
22/3
π A
R2 m/min.
(and m3/min
m2 is in metres/minute ... which is a nice check).
PS:
S: But isn't h supposed to be decreasing? I mean, why doesn't the math give dhdt < 0?
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P: Good point ... and my fault. I should have put dVdt
= -A m3/min because the volume is decreasing. Of course, I can just
assume that A is negative, then dhdt < 0.
S: Hmmm. Do I have to know the volume of a cone is 13 πr2h?
P: You can look it up, if you have time (like on an assignment). On an exam, I might give you the formula.S: Whew!P: But, later on in this course, we'll actually derive this formula.S: I can hardly wait.
Example: A rod of length 25 cm. rotates at 1000r.p.m. (revolutions per minute). Attached to the armis a second rod of length 100 cm. with one end, P,constrained to slide horizontally along the x-axis.How rapidly is P moving at any given time?
Solution: We're given the rate of change dqdt = 1000 (2π) radians per minute (since each revolution is 2π
radians). We're asked to find the rate of change: dxdt . The relation between q and x is given by the "cosine law"
for triangles: 1002 = 252 + x2 - 2 (x)(25) cos q = 252 + x2 - 50 x cos q. Now differentiate with respect to t toget:
0 = 2x dxdt - 50 (- x sin q
dqdt +
dxdt cos q). Since
dqdt is given, we can compute
dxdt for any given position (meaning
any x and q which satisfies 1002 = 252 + x2 - 50 x cos q) from dxdt = - 50
x sin q 2x - 50 cos q
dqdt . For example, when
q = 0 or π we have dxdt = 0. Similarly when q =
π2 we have
dxdt = - 25
dqdt cm/sec. ... and so on for any x and q.
PS:S: Can you check your formula with that dimensional stuff?P: Not easily, because there are numbers in there which have dimensions. It would have been better to assume the rod lengths
to be "a" and "b" (rather than 25 and 100 ) and then we'd get: dxdt = - 2a
x sin q 2x - 2a cos q
dqdt
which is dimensionally correct
since it's: (metres) (metres) (radians/second)
metres which is metres/second.
S: Whoa! I make it (metres)(radians) per second.P: Well, radians don't really have dimensions. You just have to think of any valid formula involving an angle in radians, like a
= r q (the length of arc of a circle of radius r subtending an angle q at the centre) then you'd see that q = ar which is
dimensionless ... so we don't count it. Just count length and mass and time or anything constructed with these dimensions,like force or energy or whatever.
S: I think I'll forget about this dimensional stuff.P: You may be interested to know that the problem we just solved has to do with a piston in a car engine. The crankshaft
rotates at dqdt and P slides up and down the cylinder and ...
S: Not interested.
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Example: A 10 m ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wallat 3 m/s. How quickly is the top of the ladder sliding down the wall when the bottom is 6 m from the base of thewall?
Solution: Here we know dxdt = 3 m/s and we wish to determine
dydt .
Pythagorus gives the relation between x and y, namely: x2 + y2 = 102.
Differentiating this relation with respect to t gives 2 x dxdt + 2 y
dydt = 0.
Now substitute the known quantities and solve for dydt . The known
quantities are x = 6, dxdt = 3 and y = 102 - 62 = 8 so
dydt = -
188 = -
92
m/s. (It's negative indicating that y is decreasing.)
Remember: For related rate problems, one variable, say x, has a known rate of change. Another variable, say y, is
related to x via some equation y = f(x). Then dydt =
dydx
dxdt = f '(x)
dxdt gives the rate of change of y for any
given x and dxdt . Sometimes the relation between x and y is in the form F(x,y) = C where C is some constant.
(That is, y is given implicitly in terms of x, such as x2 + y2 = 102 in the previous example). Then differentiate
implicitly to find dydt which will (usually) be in terms of x, y and
dxdt (such as
dydt = -
xy dxdt in the previous example) . To find
dydt , enough information must be given to compute
x, y and dxdt ... hence
dydt .
Example: A car moves south-west along a highway described by the curve y = x2 (where the x-axis pointseast-west and the y-axis north-south). Its headlights illuminate a fence which lies along the x-axis. Investigatethe speed with which the beam of light moves along the fence.
Solution: We need a reasonable diagram (we always need a reasonable
diagram!) The headlight beam is tangent to the curve y = x2 and will strikethe fence (i.e. will intersect the x-axis) at the x-intercept of the tangentline. So we'll pick some point of the curve (and we'll know it's on the
curve if y = x2), then we'll find the equation of the tangent line at this
point, then we'll find the x-intercept, then we'll find ddt of this x-intercept.
That's our method of attack.
Let the point on the curve be (x1,y1) where y1 = x12 ... and it's this
relation which puts the point on the curve. (Remember, we have to tell the mathematics, somehow, that the point is
on the curve and this is how we do it.) Then the tangent line equation is y - y1x - x1
= slope of tangent line = 2x1
(since dydx = 2x if y = x
2). The tangent line intersects the x-axis when y = 0 so we solve for x = x1 - y12x1
. Plug
in y1 = x12 and get the x-intercept as x = Anyway, if we take
ddt of the relation x =
x12 we get
dxdt =
12 dx1dt so
the speed with which the light travels along the fence is always half the speed with which the car moves west.There's nothing more we can do with this problem.
PS:S: Wait a minute. Nothing more? Suppose the car is moving at 100 km/hour? Then how fast is the light moving?P: That's a nice problem. Unfortunately, we have to wait until we get to "parametric equations" before we can tackle it. Remind
me when we get there and we'll come back to this problem. Wait, I'm not sure we even cover parametric equations ... whichwould really be a shame.
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S: Yeah ... a shame.
Example:
A lighthouse is located on an island 2 km from a straightshoreline. The light rotates at 3 revolutions perminute. How fast is the spot of light (at P) movingalong the shoreline when P is 4 km from thelighthouse?
Solution:
x = 2 tan q hence dxdt = 2 sec
2q dqdt
and dqdt = 3
revnsmin x 2π
radsrevn = 6π radians/minute.
When P is 4 km from the lighthouse, sec q = 42 = 2
Hence dxdt = 2 (2
2) 6π = 48π km/min.
LECTURE 10
the LINEAR (TANGENT LINE) and OTHER POLYNOMIAL APPROXIMATIONS
The TANGENT LINE APPROXIMATION
Recall that the derivative, f '(a), gives the slope of the tangent line to the curve y = f(x) at the place x = a.Further, at x = a, y = f(a) so we have a point (a,f(a)) and a slope f '(a) so we can calculate the equation of the
tangent line: y - f(a)x - a = f '(a) or (to put it into a better form)
y = f(a) + f'(a) (x - a) as if you didn't know! Is this useful? When would we really want the equation of a tangent line? In fact, the tangent
line and the curve itself are very nearly the same curve so long as we don't stray too far from the point oftangency ... and this is the basis for using the tangent line to approximate f(x) for x-values near x = a.
Note: To use y = f(a) + f '(a) (x - a) to approximate f(x), we clearly need to compute f(a) and f '(a)! Hence, weneed to choose an "a" where these can be computed with relative ease.
Example: Compute an approximate value for 47 .
Solution: Consider y = x . Then, for x = 49, y = 7. Since 47 isn't too far from 49 we find the tangent line to
f(x) = x at x = 49 and use it to compute f(47) ... approximately. The tangent line is y = f(49) + f '(49) (x - 49)
and since f '(x) = 1
2 x then f '(49) =
12(7) =
114 and our tangent line is y = 7 +
114 ( x - 49) . For x-values near x
= 49 we conclude that x ≈ 7 + 114 ( x - 49) ... that is, x and 7 +
114 ( x - 49) have approximately the same
value. In particular, 47 ≈ 7 + 114 ( 47 - 49) = 7 -
17 = 6.8571
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Above is a reasonably accurate (computer) plot of y = x and its tangent line at x = 49. A few things seemclear: (1) the tangent line is indeed a good approximation near x = 49, and (2) it gets worse as we move away
from the point of tangency, and (3) the tangent line will give an approximation to x which is too large (since
the tangent line lies above the curve). In fact, 47 = 6.8557 (to 4 decimal places). Also, (4) using the tangentline approximation (sometimes called the linear approximation because y = f(x) is approximated by a linearequation) means we move not along the curve y = f(x) but along the tangent line.
S: That looks awfully familiar! I mean, 47 ≈ 7 - 17 .
P: It should. We got that before, using the differential. In fact, y = f(a) + f ' (a) (x - a) says that y changes by Dy = y - f(a) whenx changes by Dx = x - a and if we move along the tangent line , the tangent line equation says that Dy = f ' (a) Dx and that'sthe differential! It's just a change in notation, that's all. When we used the differential we were saying that y = f(x)changes at the rate f ' (a) when x = a so a small change in x, namely Dx, would give a change in y of about f ' (a) Dx whichisn't exact because f ' (x) isn't constant, but it's good enough for small changes in x. If the velocity is 12 metres/hour then,
even if the velocity changes we'd expect to go about (12) (.1) = 1.2 metres in the next .1 hours. Nice, eh, how everythinghangs together?
Example: Show that 1 + x-12 is an approximation to x when x is near "1".
Solution: The tangent line approximation has the form f(a) + f '(a) (x-a) so we take f(x) = x and a = 1 and
get f '(x) = 1
2 x so that f(1) = 1 and f '(1) =
12 . Then f(a) + f '(a) (x-a) gives 1 +
x-12 . (For example, 1.08 ≈ 1
+ .082 = 1.04 whereas the exact root is 1.08 = 1.03923 to five decimal places.)
Example: Use the tangent line approximation to compute sin 47˚, approximately.Solution: Fortunately, 47˚ is close to 45˚ and we can compute f(x) = sin x and f '(x) = cos x at x = 45˚. The
tangent line to y = sin x is y = f(45˚) + f '(45˚) ( x - π4 ) =
1
2 +
1
2 ( x -
π4 ) . Hence, for x near
π4 ,
sin x ≈ 1
2 +
1
2 ( x -
π4 ) . Now we'd like to put x = 47˚ but we need to convert to radians. It's easier, however, to
recognize that x - π4 is just 2 degrees (i.e. 47˚ - 45˚) which, in radians, is 2
π180 =
π90 . Hence our approximate
value for sin 47˚ is 1
2 +
1
2 π90 = .7318 (while the exact value, to 4 decimal places, is .7314). Since the
tangent lies above the sine curve (at x = π/4), it's not surprising that the approximation is too large. Below is acomputer plot of the situation.
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PS:S: Two things: first, you used degrees instead of radians and you told me that ...
P: Hold on. I used radians where it mattered ... in the expression ( x - π4 ) where using degrees would be disastrous. But when I
plug in the values of sin π4 for example, I can plug in the values of sin 45˚, right? They are the same numbers, right?
S: Okay, okay, but you also said the tangent line lies above the sine curve ... so the approximation will be too large. That's easyfor you to say, but how am I supposed to know what lies above what? Am I supposed to plot the sine curve? Then I'd need acalculator and if I had a calculator why would I be interested in an approximate value for sin 47˚? I'd just punch it up on thecalculator and ...
P: Let me ask you a question ... just to see if you've been paying attention. What is it about a curve, y = f(x), that makes it liebelow its tangent line? Think about it.
S: I haven't the foggiest idea. Wait ... uh, the curve curves down. I mean, if y = f(x) is concave down then it'll be below itstangent line. Right?
P: Right on! And if it's concave up, it's above its tangent line. Now, the big question: how can you tell if a curve is concave upor down?
S: I give up.P: The second derivative! If f ' ' (x) > 0 then y = f(x) is concave up and lies above its tangent line ... and the linear (tangent line)
approximation will be too small. If f ' ' (x) < 0 then y = f(x) is concave down and lies below its tangent line ... and the linear
approximation will be too large. For f(x) = sin x, f ' ' (x) = - sin x which is negative near π4 so it's concave down and our
approximation will be too large. In the previous example we had f(x) = x = x1/2, so f ' (x) = 12 x-1/2 and f ' ' (i) = -
14 x-3/2
which is also negative near x = 49 (or near any positive x for that matter) so the tangent line will always give anapproximation which is too large.
S: I still don't understand why we'd want to use a tangent line approximation. Everybody owns a calculator. Why not use it andget the exact value?
P: Here's a nice use of the tangent line approximation:
Example: A certain amount of money is left in the bank to accumulate interest (compounded at i% per year).If you want to double your money in n years, what should the interest rate be?
Solution: If the amount of money is $A, then after n years it will have grown to $A (1 + i
100 )n. In order to
double, we need (1 + i
100 )n = 2, or, taking ln of each side, we need nln ( 1 +
i100 ) = ln 2 hence
ln ( 1 + i
100 ) = ln 2n . Now ln is a fancy function so it's not easy to solve this equation for i, but since
i100 is small
we can approximate f(i) = ln ( 1 + i
100 ) near i = 0 by its tangent line: f(i) ≈ f(0) + f '(0) i = ln 1 + 1
100 i (since
f '(i) = 1
1+i
100
1
100 = 1
100 when i = 0). Also, ln 1 = 0 so we have i
100 ≈ ln 2n . Further ln 2 ≈ .69 (roughly) so we
get i ≈ 69n . For example, to double in 10 years you'd have to get
6910 = 6.9% interest, approximately.
PS:S: How accurate is that answer? I mean 6.9% ... how accurate is it?P: The correct answer is 7.2 % (to one decimal place) ... and it's for that reason that people in the financial world use the "Rule
81
of 72" : i = 72n .
S: Well, we could have guessed that our approximation would be low because we're using the tangent line for ln ( 1 + i
100 )
and the ln curve is concave down as I recall ... I could differentiate ln ( 1 + i
100 ) twice to check, but I won't ... so the
tangent line is above the curve ... so ... hey! Why isn't the approximation too large. The tangent line is above the curve!P: This problem is different from the others. Previously we were given x and trying to evaluate y = f(x). (i.e we were given x =
47 and trying to compute 47 ... or sin 47˚). In this problem we were given y = f(x) = ln ( 1 + x
100 ) ... namely
ln 2n
... and we were trying to find x! That's different! See?S: No.
P: Let's sketch y = ln ( 1 + x
100 ) near x = 0 ... and its tangent line at x
= 0. Now, given a y-value of ln 210
, what's the x-value? The x-value on
the tangent line is smaller than the x-value on the curve ... and that'sbecause the tangent line is above the curve. See? The picture is worth...
S: That's confusing.P: Then don't think about it.
S: Just one more thing. You call the tangent line approximation a "linear" approximation. Are there other approximationswhich aren't "linear"?
P: I'm glad you asked that question ...
POLYNOMIAL APPROXIMATIONS
The thing about the approximation y = f(a) + f '(a) (x - a) is that it has the same value as f(x) and the samederivative as f(x), at x = a. In fact, if we wanted a linear approximation, meaning a straight line approximation,
y = A + Bx and we wanted y to have the value f(a) when x = a we'd need A + Ba = f(a). Also, if we wanted y to have
the same derivative at x = a then we'd want dydx = B = f '(a). This gives us two equations to solve for A and B,
namely A + Ba = f(a) and B = f '(a) . The solution is A = f(a) - a f '(a) and B = f '(a) and the linear
approximation becomes y = f(a) - a f '(a) + f '(a) x or y = f(a) + f '(a) (x - a) which is (no surprise!) the tangentline. However, when we derive it that way it's natural to ask what quadratic approximation is "best" in the sense
that it matches f(x) in value and first derivative and second derivative. So we consider y = A + Bx + Cx2 and
find the constants A, B and C by requiring that, at x = a, y = A + Ba + Ba2 = f(a) , dydx = B + 2Ca = f '(a)
and d2y
dx2 = 2C = f ' ' (a) . After solving these 3 equations in 3 unknowns and substituting we get the quadratic
approximation:
y = f(a) + f'(a) (x - a) + 12 f''(a) (x-a)
2
Actually, the procedure is easier if we started off by assuming a quadratic approximation in the form:
y = a0 + a1 (x - a) + a2 (x - a)2 (which is still a quadratic ... because the highest power of x is x2 ... but in a form
which simplifies the calculations). Then, at x = a, we have y = a0 and dydx = a1 and
d2y
dx2 = 2 a2 and
equating these to f(a), f '(a) and f ' ' (a) respectively gives, immediately, a0 = f(a), a1 = f '(a) and a2 = 12 f ' ' (a),
hence the required quadratic approximation, as above.Note that the quadratic approximation is, in some sense, the "best" parabolic approximation to the curvewhereas the tangent line is the "best" straight line approximation ... at least at x = a. We could also determine the
82
"best" cubic approximation or quartic approximation, etc. etc. In fact, since a cubic y = A + Bx + Cx2 + Dx3 ...
or, better, y = a0 + a1 (x - a) + a2 (x - a)2 + a3 (x - a)
3 ... has precisely 4 constants, we can insist upon 4conditions (yielding 4 equations to solve for the 4 unknown constants). The 4 conditions are that, at x = a
(meaning any specified value of x), the values of y, dydx ,
d2y
dx2 and
d3y
dx3 must agree with f(a), f '(a), f ' ' (a) and
f ' ' ' (a). We'd get the same values for a0, a1 and a2 as we did for the quadratic approximation ... and a3 would be
16 f ' ' ' (a).
As with the linear approximation, we usually pick "a" at a place where f(a), f '(a), f ' ' (a) and f ' ' ' (a) aren'tdifficult to compute.
Example: Determine a quadratic approximation to x at x = 49.
Solution: Note that a = 49 is someplace where we can easily evaluate f(x) = x1/2, f ' (x) = 12 x
-1/2 and
f ' ' (x) = - 14 x
-3/2. We get f(49) = 17 , f '(49) = -
114 , f ' ' (49) =
11302 so our quadratic approximation (at x = 49)
becomes y = 7 + 114 (x - 49) -
12
11302 (x - 49)
2 = 7 + 114 (x - 49) -
12604 (x - 49)
2 . To see how good it is, we
compute an approximation to 47 as 7 + 114 (47 - 49) -
12604 (47 - 49)
2 = 6.85561 (to 5 decimal places). The
exact value (also to 5 decimal places) is 47 = 6.85566 and the linear, tangent line approximation gives6.85714 (to 5 decimal places). This is illustrated in the graph below:
S: The tangent line approximation is just the first two terms of the quadratic approximation. Is that always the case? I mean, toget the quadratic approximation we just added another term. To get the cubic approximation, do we add yet another term?
P: Sure, and remember the format of these polynomial approximations:
POLYNOMIAL APPROXIMATIONS
linear: y = f(a) + f'(a) (x - a)
quadratic: y = f(a) + f'(a) (x - a) + 12 f''(a) (x-a)
2
cubic: y = f(a) + f'(a) (x - a) + 12 f''(a) (x-a)
2 + 16 f'''(a) (x-a)
3
S: Hold on. I see how it goes ... the fourth degree approximation would get another term ... uh, something with
f ''''(a) (x-a)4 ... but what's the number out front? There's 12 then comes
16 ... then what?
P: Write y = a0 + a1 (x - a) + a2 (x - a)2 + a3 (x - a)
3 + a4 (x - a)4. Then differentiate four times and put x = a and notice that
everything differentiates to zero except the term a4 (x - a)4 which becomes (4)(3)(2)(1) a4 so
d4y
dx4 = 4! a4 and this must
83
equal f ''''(a) hence a4 = 14! f ''''(a). The number out front, as you put it, is
14! =
124
(just like 16 is really
13! and
12 is really
12! ) . Now it's easy to see what's the polynomial approximation of degree 5 or 6
or whatever.
Example: Determine an approximate value for sin 47˚ using a quadratic approximation.Solution: We consider f(x) = sin x and determine the quadratic approximation at x = 45˚ (since we know all
about the sine function and its derivatives at this x = a). We have f(x) = sin x = 1
2 and f '(x) = cos x =
1
2 and
f ''(x) = - sin x = - 1
2 (all evaluated at x = 45˚, or x =
π4 in radians) . Our quadratic approximation is then
y = f(a) + f '(a) (x - a) + 12 f ''(a) (x-a)
2 with a = π4 and this gives: y =
1
2 +
1
2 (x -
π4 ) -
1
2 (x -
π4 )
2 (where we
were careful to substitute a = 45˚ in radians in (x - a) and (x -a)2 etc.) Now, to get an approximation to sin 47˚
we need to express x - π4 in radians; that is, 2˚ in radians which is 2
π180 =
π90 so finally we get:
sin 47˚ ≈ 1
2 +
1
2
π
90 - 1
2
π
90 2 = .731359 (to 6 decimal places). To 6 dec. places, the exact value is
.731354 and the linear approximation, 1
2 +
1
2
π
90 , is .731789 (so we can appreciate the improvement in
going to the quadratic approximation, and, as you'd expect, the cubic approximation is even better).
PS:P: Did you notice anything interesting about the formulas for the linear and quadratic approximations?S: Nope.
P: Let me write them out again: y = f(a) + f ' (a) (x - a) versus y = f(a) + f ' (a) (x - a) + 12 f ' ' (a) (x - a)2. Remember? If the
curve y = f(x) lies below its tangent line, at x = a, then the linear approximation is too large ... and we can predict that bylooking at the sign of the second derivative, f ' ' (a). Now do you see?
S: Uh ... not really.P: When f ' ' (a) < 0 then y = f(x) is concave down and the curve lies below its tangent line so the linear approximation is too
large ... so we should really subtract something from the linear approximation ... and the quadratic approximation does just
that! It adds 12 f ' ' (a) (x-a)2 to the linear approximation which, if f ' ' (a) < 0, really means it's subtracting something. Neat,
eh?S: Aaah ... mathematics is wonderful. Let's do some more ... something really useful ... if there is anything useful about this
stuff ... my calculator can do all this ... I don't know why we're studying this ...
Example: Solve the equation ln (1 + i
100 ) = ln 210 for i, using a quadratic approximation for the logarithm.
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Solution: We have f(i) =ln (1 + i
100 ) so f '( i ) = 1
1 + i
100
1
100 = 1
100 + i and f ' ' ( i ) = - 1
(100 + i)2
and, evaluating at i = 0 (we choose this value since it's easy to compute f(0), f '(0) , e tc . ) we have f(0) = ln 1 =
0, f '(0) = 1
100 and f ' ' (0) = - 1
1002 hence our quadratic approximation is y = f(0) + f '(0) (i - 0) +
12 f ' ' (0)
(i - 0)2 = 1
100 i + 12
-1
1002 i2 =
i100 -
12
i
100 2. Then we solve
i100 -
12
i
100 2 =
ln 210 which is a quadratic
equation to solve for i. (This is no surprise since we're using a quadratic approximation for f(x)!) To solve, it'seasier to let
x = i
100 then solve for x. We rearrange the equation, rewriting it in the form: x2 - 2 x + ln 25 = 0 and use the magic
formula for the roots of a quadratic: -b ± b2 - 4ac
2a to get x = 1 ± 1 - ln 25 = 1.9281 or .0719 and we pick
the smaller root and conclude that i
100 ≈ .0719 so i = 7.19% (which gives the "Rule of 72").
If we look at the graph of y = ln (1 + x) and thequadratic (i.e. parabolic) approximation, we seethat it's excellent near x = 0 and becomes worse aswe move away from the point of approximation.Further, the quadratic approximation is too smallfor x > 0 and too large for x < 0 (although it's hardto tell from the diagram).Also, the curve y = ln (1 + x) has a verticalasymptote at x = -1 (since ln (1 + x) -> -∞) and we
wouldn't expect the parabola y = x - x2
2 to be any
good as an approximation anywhere near x = -1.Even a cubic or quartic approximation won't have avertical asymptote at x = -1 hence can't providegood approximations there.
PS:S: Why did you discard the root 1.9281?
P: Instead of solving ln (1 + x) = ln 210
which has
exactly one solution for x (hence for i) ... because it's an increasing function ... we solved x - x2
2 =
ln 210
and this equation has
two solutions (as most quadratic equations do!). In fact, looking at the graph above we see that, although y = ln (1 + x) has
the value ln 210
only once, y = x - x2
2 has this value twice ... since the parabola goes up then comes back down. It's the first
one that we want ... the one nearest to the point of approximation ... the one nearest x = 0. One thing I should mention:although the graphs were plotted on a computer and show the beautiful quadratic approximation, the horizontal dotted line
is NOT y = ln 210
≈ .069 (since I wanted to illustrate the two intersections with the quadratic, and it was clearer with a line
something like y = .4). Anyway, as you can plainly see, the picture is worth ... well ... you know.
85
Problem: Verify each of the polynomial approximations illustrated above.
Note: For the first, f(x) = ln x, the polynomial is constructed about x = 1. Why can't we find a polynomial (of thetype we've been constructing) about x = 0?
PS:S: You've been calling these the "best" quadratic and the "best" cubic approximations, etc. Aren't they the "best"? And if so,
why do you put "best" in quotes. It's as though you're not convinced ... or not sure, or something.
P: Okay, let's consider something simple, say, the "best" linear approximation to f(x) = x3 where we'll take the approximationabout x = 0. Then f(0) = 0 and f ' (0) = 0 as well and our "best" linear approximation isy = f(0) + f ' (0) x = 0 (using f(a) + f ' (a) (x - a) with a = 0). Hence our "best" straight-line approximation is the x-axis itself!If we look at the graph, it's easy to see that when we require our "best" line to match the value f(0) and the derivative f ' (0), itgives a pretty poor approximation. In fact, y = x looks quite a bit better (as a straight-line, "linear" approximation). On theinterval -1 ≤ x ≤ 1, for example, our "best" (tangent line) approximation, y = 0, has a maximum error of "1" whereas the
maximum error using the approximation y = x is only ... uh, let's see ... the error is the difference between y = x and y = x3
and that's | x - x3| and, on 0≤x≤1 we can delete the absolute value sign because x - x3 ≥ 0, so this error has a maximumvalue, on 0 ≤ x ≤ 1 either at a critical point in 0 < x < 1, or at the end-points x = 0 or x = 1. The critical points are whereddx
(x - x3) = 1 - 3x2 = 0, meaning x = 1
3 (on 0 < x < 1) so the maximum is
1
3 -
1
( 3 )3 ≈ .38 which is better than "1".
See?S: See what?
86
P: The so-called "best" linear approximation for f(x) = x3 (using themethod we've been describing ... that is, matching the value andderivatives of f(x) at x = 0, and so on) doesn't do as well as y = x. Infact, there are even better approximations than y = x . Of course, it alldepends upon what one means by "better", doesn't it?
S: If you say so.P: You might consider the line y = mx and try to vary the slope "m" and
see which gives the smallest maximum error. It's another meaning forthe word "best" ... and it's fun.
S: If you say so.
LECTURE 11
NEWTON'S METHOD for finding roots of f(x) = 0
In earlier lectures we've run into a variety of problems which require finding roots of equations of the form:
f(x) = 0. For example, we've considered 99 x4 - 594 x3 + 1290 x2 + 6x - 9 = 0, and 47 which is a root of the
equation x2 - 47 = 0, and ln (1 + x) = ln 210 which can also be written in the form f(x) = 0 with
f(x) = ln (1 + x) - ln 210 . Now we use the linear tangent line approximation in a special way, called Newton's Method,
to find such roots to as many decimal places as we wish.First note that the curve y = f(x) crosses the x-axis at an x-valuewhich satisfies f(x) = 0, so finding a root is the same as findingan x-intercept for such a curve. Suppose we sketch, or plot, y =f(x) and find that it changes sign between, say x = a and x = b.That is, f(a) has a different sign than f(b); for example, in thediagram, f(a) < 0 and f(b) > 0. Then there is a root lying in a < x< b. Now we pick any number x1 in a ≤ x ≤ b (it could be either
"a" or "b", or a number in between as shown in the diagram) andconstruct the tangent line to y = f(x) at x = x1. This tangent line
will intersect the x-axis at some x-value, say x2, and this will be
closer to the root
of f(x) = 0 than was x1. In other words, if x1 is an approximation to the root, then x2 is a better approximation.
So what's x2?
The tangent line at x = x1 is: y = f(x1) + f '(x1) (x - x1) and this crosses the x-axis at x = x2 where
y = 0 = f(x1) + f '(x1) (x2 - x1) and, solving, we get: x2 = x1 - f(x1)
f '(x1) . If x2 is better than x1 (as an
approximation), then we can repeat the procedure, finding the tangent line at x = x2 and determining x3, where it
crosses the x-axis. There's no need to repeat the calculation; it's similar to that found above: x3 = x2 - f(x2)
f '(x2) .
We can now repeatedly use the same prescription, generating a sequence of number x2, x3, x4, ..., xn, ... which
get closer and closer to the root of f(x) = 0 ... or, to use a more apt notation:
limn->∞
xn = c
87
From the diagram at the left you can watch the sequence ofapproximations marching toward the root of f(x) = 0, at x = c.Just remember that x1 is some initial approximation which we
generate by guessing or by plotting or by divine insight. The restof the approximations come from Newton's formula:
xn+1 = xn - f(xn)f'(xn)
Example: Compute 47 , using Newton's Method.
Solution: We invent the function f(x) = x2 - 47 so that the positive root of f(x) = 0 will give us 47 . Note
that we don't invent the function f(x) = x - 47 (even though f(x) = 0 will certainly have the root x = 47 )
since we can't evaluate f(x) if we don't know 47 !! (We want to "evaluate" f(x) and f '(x) using only addition,subtraction, multiplication and division; think of doing all this on a $3.95 calculator which doesn't have asquare root button.) Then, to compute the Newton iterates (that's what the x1, x2, etc. are called) we guess at a
root, say x1 = 7 (which seems a reasonable guess) and plug this x-value into x - f(x)f '(x) = x -
x2 - 472x =
12
x +
47x
where we've simplified the expression somewhat. Then
x2 = 12
x1 +
47x1
= 12 (7 +
477 ) = 6.857142858 and now we plug this x2 into the same expression giving
x3 = 12
x2 +
47x2
= 12 (6.857142858 +
476.857142858 ) = 6.855654762 and we now we plug this x3 into the same
expression giving
x4 = 12
x3 +
47x3
= 12 (6.855654762 +
476.855654762 ) = 6.855654601 and we now we plug this x4 into the
same expression giving
x5 = 12
x4 +
47x4
= 12 (6.855654601 +
476.855654601 ) = 6.855654601 and if we continue we'll get x6 =
6.855654601 as well. We're finished. To 9 decimal places, that's 47 .Of course, if we carried more decimal places we could go and on, getting better and better approximations ...and many more decimal places ... and the error would approach zero. In fact, it's very instructive to actuallywatch the error go to zero. To do this we'll begin again with x1 = 7 but now we'll carry 75 digits.
The numbers which follow were computed using : a computer algebra system from U. of Waterloo.
(My $4.95 calculator can't do 75 digits). When we have need of lots of digits, we'll let do it.
In what follows, we first ask for 75 digits (Digits:=75;) then ask to evalf (meaning evaluate as afloating point, or decimal, number) the exact square root (and we'll call it "root"), then we'll define
y = (x + 1/x)/2 and we'll "iterate", computing x1, x2, etc. and each time we'll compute an error = x2 - root, etc._______________________________________________________________________________• Digits:=75; Digits := 75_______________________________________________________________________________• root:=evalf(sqrt(47)); root :=
6.85565460040104412493587144908484896046064346100132627548510818567851711514_______________________________________________________________________________• y:=.5*(x+47/x); y := .5 x + 23.5 1/x_______________________________________________________________________________• x2:=subs(x=7,y); x2 :=
6.85714285714285714285714285714285714285714285714285714285714285714285714286_______________________________________________________________________________• error:=x2-root; error :=
.00148825674181301792127140805800818239649939614153086737203467146434002772_______________________________________________________________________________
88
• x3:=subs(x=x2,y); x3 :=
6.85565476190476190476190476190476190476190476190476190476190476190476190476_______________________________________________________________________________• error:=x3-root;
-6 error := .16150371777982603331281991294430126130090343562927679657622624478962*10_______________________________________________________________________________• x4:=subs(x=x3,y); x4 :=
6.85565460040104602726699535902922054430628339896844346594726451466195334222_______________________________________________________________________________• error:=x4-root; -14 error := .190233112390994437158384563993796711719046215632898343622708*10_______________________________________________________________________________• x5:=subs(x=x4,y); x5 :=
6.85565460040104412493587144908511289322242413572750032746808991073007441445_______________________________________________________________________________• error:=x5-root; -30 error := .26393276178067472617405198298172505155729931*10_______________________________________________________________________________• x6:=subs(x=x5,y); x6 :=
6.85565460040104412493587144908484896046064346100132627548510819075903143518_______________________________________________________________________________• error:=x6-root; -62 error := .508051432004*10_______________________________________________________________________________• x7:=subs(x=x6,y); x7 :=
6.85565460040104412493587144908484896046064346100132627548510818567851711514_______________________________________________________________________________• error:=x7-root; error := 0_______________________________________________________________________________
Did you see? The error goes to zero! (Well ... at least "zero", to 75 digits). Remarkable! The errors are
something like 10-3, 10-7, 10-15, 10-31, 10-62 then something less than 10-75 (in fact, probably about 10-120).In fact, it's this remarkable rapidity with which Newton's Method gives roots that endears it to many. Once aniterate finds itself near a root, the remaining iterates march to the root with unerring accuracy. Each successiveerror is roughly the square of the preceding error!
Example: Compute a root of 99 x4 - 594 x3 + 1290 x2 + 6x - 9 = 0 (to 5 decimal places).Solution:
This problem arose (earlier) in connection with finding the path,from P to Q, which takes the minimum time. Point Q is on landand P is in water and swimming speed is 1/10 of running speed.
For f(x) = 99 x4 - 594 x3 + 1290 x2 + 6x - 9 we have theiteration scheme based upon:
x - f(x)f '(x) = x -
99x4 - 594x3 + 1290x2 +6x - 9
396x3 - 1782x2+2580x + 6 . Come to think of it,
maybe I'll let do it. We'll define y = f(x), ask
to compute y' , construct the iteration equation x - y/y' , substitute various values of x into y ... until wefind two values where y changes sign, then we'll pick one of these two x-values as x1 ... and iterate, until the
iterates are identical to, say, 6 digits. Note: in the following excerpt from a session with , thecurious command "evalf" means evaluate as a floating point (i.e. decimal) number.
_________________________________________________________________________________
89
• Digits:=6; Digits := 6_______________________________________________________________________________• y:=99*x^4 - 594*x^3 + 1290*x^2 + 6*x - 9; 4 3 2 y := 99 x - 594 x + 1290 x + 6 x - 9_______________________________________________________________________________• Dy:=diff(y,x); 3 2 Dy := 396 x - 1782 x + 2580 x + 6_______________________________________________________________________________• iterate:=x - y/Dy; 4 3 2 99 x - 594 x + 1290 x + 6 x - 9 iterate := x - ------------------------------------ 3 2 396 x - 1782 x + 2580 x + 6_______________________________________________________________________________• subs(x=0,y); -9_______________________________________________________________________________• subs(x=1,y); 792_______________________________________________________________________________• x1:=0; x1 := 0_______________________________________________________________________________• x2:=evalf(subs(x=x1,iterate)); x2 := 1.50000_______________________________________________________________________________• x3:=evalf(subs(x=x2,iterate)); x3 := .33712_______________________________________________________________________________• x4:=evalf(subs(x=x3,iterate)); x4 := .165492_______________________________________________________________________________• x5:=evalf(subs(x=x4,iterate)); x5 := .101483_______________________________________________________________________________• x6:=evalf(subs(x=x5,iterate)); x6 := .0843389_______________________________________________________________________________• x7:=evalf(subs(x=x6,iterate)); x7 := .0827736_______________________________________________________________________________• x8:=evalf(subs(x=x7,iterate)); x8 := .0827600_______________________________________________________________________________• x9:=evalf(subs(x=x8,iterate)); x9 := .0827601_______________________________________________________________________________• x10:=evalf(subs(x=x9,iterate)); x10 := .0827601_______________________________________________________________________________
We conclude that a root of f(x) = 99 x4 - 594 x3 + 1290 x2 + 6x - 9 = 0 (to 5 decimal places) is .08276 ...
PS:S: Wait! How come it took so long ... and you only asked for 6 digits?P: Remember, we're approximating y = f(x) by its tangent line and actually finding where this tangent line intersects the x-axis.
Maybe the tangent line isn't a good approximation to y = f(x), for this particular f(x).S: How could we check that?P: Well, if the slope of f(x) doesn't change too rapidly near x = .0827601, then we'd expect the tangent line to be pretty good.
After all, the tangent line has a constant slope but y = f(x) doesn't. When we wanted to compute y = x2 - 47 near x = 47 ,
the Newton iterates quickly got us to a root. For y = x2 - 47 , dydx
= 2x and d2y
dx2 = 2 which isn't so large so
dydx
isn't changing
90
too rapidly. Let's check y = 99 x4 - 594 x3 + 1290 x2 + 6x - 9 near the root we just found. gives:
_______________________________________________________________________________• y:=99*x^4 - 594*x^3 + 1290*x^2 + 6*x - 9; 4 3 2 y := 99 x - 594 x + 1290 x + 6 x - 9_______________________________________________________________________________• Dy:=diff(y,x); 3 2 Dy := 396 x - 1782 x + 2580 x + 6_______________________________________________________________________________• D2y:=diff(Dy,x); 2 D2y := 1188 x - 3564 x + 2580_______________________________________________________________________________• evalf(subs(x=.0827601,D2y)); 2293.179894_______________________________________________________________________________
so the second derivative is quite large (and the curve bends rapidly away from its tangent line ... so the tangent line isn't avery good approximation near x = .0827601).
S: Are there any other roots of f(x) = 99 x4 - 594 x3 + 1290 x2 + 6x - 9 = 0?P: I don't know. Shall we plot it and see? We can just evaluate f(x) for a whole bunch of x-values and see if it changes sign
anywhere, then we can use Newton's method again, starting with an initial guess, x1, nearby and ..
S: Let's forget it ... except ... I almost hate to ask, but will I ever really have to find the root of 47 or where to swim to get to thecottage or ...
P: Okay, that's a fair question: "When would anyone ever need Newton's Method, outside of a calculus course?"
Example: You invest $10,000 in a mutual fund, then, 5 months later you put an additional $15,000 into thefund, then, 3 months later put in an additional $5,000. At the end of a year, your investments (totalling $30,000)have grown to $31,470. What is the annual rate of return from this mutual fund?
Solution: Let the monthly rate of return be i . For example, if i =.01 it means a 1% return on your investment
per month so each dollar will grow to (1.01)n after n months. The first $10K invested has been in the fund for 12
months, so will grow to 10(1+i)12 (measured in kilobucks). The next $15K grows to 15(1+i)7, having been in
the fund for 7 months. The last $5 will grow to 5(1+i)4, having been in the fund for 4 months. The total value of
your investments, after 12 months, is 10(1+i)12 + 15(1+i)7 + 5(1+i)4 = 31.47 kilobucks. We must solve thisequation for i.
We let 1+i = x and rewrite the equation in the standard format: f(x) = 10x12 + 15x7 + 5x4 - 31.47 = 0 and use
the iteration scheme based upon x - f(x)f '(x) = x -
10x12 + 15x7 + 5x4 - 31.47
120x11 + 105x6+20x3 . If we guess at a monthly return of
1% (we need an initial guess!), then we can use i = .01 hence x1 = 1+i = 1.01 is our initial guess. Iterating, we
get ...
S: Wait! How many iterations will it take? Guess!P: Well, it depends upon how rapidly the slope of f(x) changes and that depends upon f ' ' (x) and
f ' ' (x) = 1320x10+630x5+60x2 and, for x = 1.01 that gives ... did you bring your calculator?S: Just stick in x = 1, that's close enough ... and that gives f ' '= 1320+630+60 = 2010 and that's pretty big, right?P: You're getting very clever. Yes, it's large, so I'd say about 6 or 7 iterations. Of course, it depends upon how close x1 is to the
root of f(x) = 0. Anyway, let's proceed ... in fact, let's use a computer spread sheet where I've programmed:
x2 = x1 - 10x1
12 + 15x17 + 5x1
4 - 31.47
120x111 + 105x1
6+20x13
, x3 = x2 - 10x2
12 + 15x27 + 5x2
4 - 31.47
120x211 + 105x2
6+20x23
and so on.
In what follows, I type in the numbers shown in boldface and the spreadsheet does the rest. I also have the spreadsheet
compute (for each iteration x1, x2, etc.) the equivalent Annual Rate, as a percentage , namely 100(x12 - 1):
91
S: Aha! You only needed two iterations!
P: Amazing! Now you can see how clever Newton is ... uh, was. However, we can also estimate the root of f(x) = 0 by using aquadratic approximation for f(x). Want to try it?
S: Why not.
We'll use a quadratic approximation for f(x) =10x12 + 15x7 + 5x4 - 31.47, about x = 1 (since we expect x = 1+ito be near 1).
We have f(x) ≈ f(1) + f '(1) (x-1) + f ' ' (1) (x-1)2
2 = -1.47 + 245 (x-1) + 1005 (x-1)2. Instead of solving
f(x) = 0, we solve the quadratic equation 1005 (x-1)2 + 245 (x-1) - 1.47 = 0 or putting x - 1 = i, we solve
1005 i2 + 245 i - 1.47 = 0. Using the world famous formula i = -b± b2-4ac
2a gives i = .00586 where we took the
positive square root because we clearly want i positive! This compares quite favorably with the value obtainedvia Newton's method.
S: Favorably? It's right on the button!P: Well, to 5 decimal places. To 7 decimal places the exact root (using Newton's method) is i = .0058570 whereas the quadratic
approximation gives i = .0058592 which is pretty good, eh what?S: Yeah. One other thing ... does Newton's method always work?P: I'm glad you asked that question:
DIFFICULTIES WITH NEWTON'S METHOD:If we begin our iterations with x1 far from a root of f(x) = 0, we can't guarantee that the iterates x2, x3, etc. will
march to the root. Since we have a nice geometrical picture of what the method is doing (i.e. repeatedly movingalong tangent lines to the x-axis) we can see what happens in various cases:
Suppose f(x) has two roots, x = p and x = q, and webegin at x = x1, too far to the right of q (which, we
assume, is the root we want). It's possible that thetangent line at x = x1 will intersect the x-axis nearer
to x = p and the iterates will have a limit of p(instead of q). That's not too bad ... at least we get aroot.
Worse still, we might pick an x1 where the tangent
line is horizontal ... then it never intersects the x-
axis! (In this case, x2 = x1 - f(x1)
f '(x1) involves a
division by f '(x1) which is zero.)
92
Also, we might pick x1 so that the tangent line
intersects the x-axis (at x2) too far to the right of q
and the next tangent line takes us a mile or twoalong the negative x-axis ... and who knows whatf(x) looks like there?
Also, we might find that x3 is identical to x1 and
the iterates just repeat: x1, x2, x1, x2, etc. etc. not
having any limiting value at all.
As you might imagine, there are other weird thingsthat can happen.
The moral? Pick a reasonable value for x1 before you start iterating!
Before we leave Newton's method, let's look once again at the iterative procedure for finding a square root. To
find N we invent f(x) = x2 - N and, to solve f(x) = 0, we iterate using x - f(x)f '(x) = x -
x2 - N2x =
12
x +
Nx and
some reasonable first iterate, x1. (i.e. we use xn+1 = 12
xn +
Nxn for n = 1, 2, 3, 4, ... etc.). But look at how
clever this scheme is! If x1 is smaller than N , then Nx1 will be larger than N so Newton's method picks, as
x2, the average of x1 and Nx1 ! In fact, at each stage of our iterations, xn and
Nxn will be on either side of
N (one larger, one smaller) so we actually have the root in some interval which (hopefully) gets smaller andsmaller. Very nice. This technique makes such sense that it's used by people who have never heard of Newton!
Examples: (a) Solve x3 - x2 + x + 22 = 0.(b) Solve x ln x = 6
Solutions: (a) Using xn+1 = xn - f(xn)
f '(xn) with f(x) = x3-x2+x+22
we get: xn+1 = xn - xn
3-xn2+xn+22
3xn2-2xn+1
. The only root is
slightlylarger than 3, so we use x1=3 and get x2=3.045789, x3=3.044724,
x4=3.044723, x5=3.044723 and we conclude that the root is 3.04472 to
five dec. places.
(b) With f(x) = x ln x - 6 we get:
xn+1 = xn - xn ln xn - 6
ln xn + 1 . There is one root between 4 and 5. If we use x1=4.3
we get x2=4.189351, x3=4.188760, x4=4.188760 and we
conclude that the root is 4.18876 to five dec. places.
93
LECTURE 12
L'HÔPITAL'S RULE for evaluating limits of the form 00
In an earlier lecture we investigated various LIMIT RULES (such as lim (f + g) = lim(f) + lim(g) and so on)which allowed us to avoid computing limits by resorting to the definition of "limit". Often these rules were of no
value, such as would be the case for the rule:
limx->a
fg =
limx->a
(f)
limx->a
(g)
, in the case where
limx->a
(g) = 0. However, if
limx->a
f = L (a number different from zero) and
limx->a
g = 0 we agreed to use the notation
limx->a
fg = ∞ (perhaps
+∞ or -∞ depending upon the sign of L and whether
limx->a
g(x) = 0 from the right or from the left). One of the
most irritating limits is the case where
limx->a
g(x) = 0 AND
limx->a
f = 0 ... then we say that the ratio f(x)g(x) has the
indeterminate form 00 . It's this form which we want to consider now.
To begin, let's consider f(x) = x2 - 4 and g(x) = x3 - 8 and
limx->2
f(x)g(x) which has the
00 form. (I know we can
factor both numerator and denominator, cancel the factor (x - 2) and then use the limit rule ... but we won't
because we want to generate another technique.) We rewrite the ratio in the form
limx->2
f(x) - f(2)
x - 2
g(x) - g(2)x - 2
which
doesn't change anything because f(2) = 0 and g(2) = 0 and we've divided both numerator and denominator by (x
- 2) which isn't zero if x->2 (because we want x near 2, but not equal to 2). Now note that
limx->2
f(x) - f(2)
x - 2 =
f '(2) and
limx->2
g(x) - g(2)
x - 2 = g '(2) so NOW the numerator and denominator have limits different from zero (i.e. we've
eliminated the dreaded 00 form), and we get
limx->2
f(x) - f(2)
x - 2
g(x) - g(2)x - 2
= f '(2)g'(2) = [
2x
3x2 ]x=2 =
13 , where the
convenient notation [ ]x=2
means "evaluated at x = 2".
This prescription for evaluating limits of the indeterminate form 00 is called:
94
l'HÔPITAL'S RULE for 00
If
limx->a
f(x) = 0 and
limx->a
g(x) = 0 then
limx->a
f(x)g(x) =
limx->a
f'(x)g'(x) provided this limit exists
Of course, in order for l'Hopital's rule* to be valid, the functions f(x) and g(x) must have derivatives nearx = a (else the limit of their ratio has no meaning).
Note: The 00 form is precisely the form of the difference ratio that defines the derivative:
limx->a
f(x) - f(a)
x - a .
If we apply l'Hopital's rule we get
limx->a
f(x) - f(a)
x - a =
limx->a
ddx( )f(x) - f(a)
ddx(x - a)
=
limx->a
f ' (x)1 = f '(a) as expected!
Examples: Evaluate the following limits
(a)
limx->0
1 - ex
x (b)
limx->0
1 - cos x
x2 (c)
limx->∞
x2e-x
Solutions:
(a)
limx->0
1 - ex
x =
limx->0
- ex
1 = -1 (where we differentiated numerator and denominator).
(b)
limx->0
1 - cos x
x2 =
limx->0
sin x2x =
limx->0
cos x2 =
12 where the first application of l'Hopital's rule yielded a limit
still in the form 00 (namely =
limx->0
sin x2x ) , so we used l'Hopital again!
(c)
limx->∞
x2e-x =
limx->∞
e-x
1
x2
(putting it into the requisite 00 form) =
limx->∞
- e-x
-2
x3
(applying l'Hopital)
=
limx->∞
12 x
3 e-x (rearranging a bit) and we wind up with an expression which is worse than the one we
started with! If we continue in this manner, the limits will NOT get easier to evaluate ... they'll get worse ...
and that brings us to the second form, ∞∞ , which, miraculously, also succumbs to l'Hopital's rule:
l'HÔPITAL'S RULE for ∞∞
* The wealthy Marquis de l'Ho ^pital had one of the famous Bernoullis as his math teacher. Apparently, in 1955, a
letter was discovered which indicated that it was Bernoulli and not l'Ho ^pital who was the author of this "rule" ...
l'Ho ^pital simply paid Bernoulli so that the Marquis could claim it as his own! Nevertheless, l'Ho ^pital did publishthe very first calculus book.
95
If
limx->a
f(x) = ∞ and
limx->a
g(x) = ∞ then
limx->a
f(x)g(x) =
limx->a
f'(x)g'(x) provided this limit exists
As you might imagine, the functions must actually have derivatives in order to use this rule.
Now we write
limx->∞
x2e-x =
limx->∞
x2
ex (which now has the
∞∞ form) =
limx->∞
2x
ex (applying l'Hopital) =
limx->∞
2
ex (applying l'Hopital one last time) = 0.
P: Do you recognize
limx->∞
x2e-x = 0?
S: Nope.
P: We talked about it when I mentioned the explosive growth of ex. In fact, I think I said that x1000
ex has a limit of zero
because no matter how hard x1000 tries to drag the fraction to ∞, the ex in the denominator ...S: ... drags it to zero. Yeah, I remember now.
P: Well, now we can see that:
limx->∞
x1000
ex =
limx->∞
1000x999
ex =
limx->∞
1000(999)x998
ex
=
limx->∞
1000(999)(998)x997
ex = ... etc. etc. =
limx->∞
1000(999)(998) ... (3)(2)(1)
ex =
limx->∞
1000!
ex = 0.
See? We keep differentiating numerator and denominator until we no longer have the form ∞∞ , and ex just waits patiently
until x1000 is differentiated to a constant, then it drags the fraction to zero.S: Are you going to prove l'Hopital's rule?P: No, just trust me; it works.
Interpretation of a Limiting Value:
Having evaluated a limit, say
limx->a
f(x) = L, then we know that | f(x) - L | is very small when x is close to "a". In
fact, that sometimes says something quite interesting about f(x) in cases where f(x) = p(x)q(x) , a ratio of functions
(which is what we've been considering here). If p(x)q(x) -> L, then
p(x)q(x) is "close to" L (when x is near "a").
Consider, for example, the by-now-familiar
limx->0
sin xx = 1. It says that
sin xx is very nearly "1" when x is close
to zero, so y = sin x and y = x have nearly the same values for small values of x. (We pointed out, earlier, thatchoosing x = .0123 we find that sin (.0123) = .01229969). In a sense, we're comparing the values of sin x and x.
We can do the same for
limx->0
1 - ex
x = -1 so that the values of 1 - ex are roughly the negative of the values of x
(when x is small, of course). That is: 1 - ex ≈ -x for small x, hence ex ≈ 1 + x.
Similarly
limx->0
1 - cos x
x2 =
12 tells us that 1 - cos x ≈
x2
2 hence cos x ≈ 1 - x2
2 when x is small.
Similarly, we might ask: If sin x is "close to" x (when x is small), then how about sin x - x? What's it "close to"?
We might try to compare sin x - x with, say, x2 by evaluating
limx->0
sin x - x
x2 =
limx->0
cos x - 1
2x
96
=
limx->0
- sin x2 = 0, which says that sin x - x is very much smaller than x2 (when x is very small) since the ratio of sin
x - x and x2 approaches zero. Let's then compare sin x - x with something smaller than x2 ... say x3 (which is
certainly smaller than x2 when x is small). We have
limx->0
sin x - x
x3 =
limx->0
cos x - 1
3x2 =
limx->0
- sin x6x =
- cos x6 = -
16 (where we've had to apply l'Hopital's rule three times to get rid of the
00 form) . We now have that sin x - x is
"very close" to - 16 x
3 (for small x), or, sin x ≈ x - x3
6 and that's the cubic approximation we considered earlier.
In fact, 1 - x2
2 is the quadratic approximation to cos x (at x = 0) and 1 + x is the linear (tangent line)
approximation to ex (at x = 0).
S: Why are you always writing this is "close to" that? Why the quotes? Are they close, or aren't they?P: Good question! Now I have one for you: how do you measure "closeness". I mean, if I gave you two numbers, what would
you do to see if one was "close to" the other?S: I'd just look at them ... what else?P: No. Give me a prescription which I could place in a "Manual on Closeness" ... some algorithm or procedure which anyone
could follow. You can't say "when you have two numbers you just look at them". I see you don't know what I mean. Okay, Ihave in my pocket two numbers. Their difference is .00001 and I'd like to know it they're "close" in value.
S: Of course they are.P: The two numbers happen to be .00002 and .00001, the first being 100% larger than the second, yet their difference is
.00001, so do you still think these two numbers are "close"?S: Huh?P: The distance to the sun is 150,000,000 km and to an asteroid is 149,500,000 km. Are these two distances "close"?S: I'd say so.P: Yet their difference is 500,000 which is a pretty large. Funny eh? .00002 is NOT "close to" .00001, yet their difference is
only .00001, while 150,000,000 is "close to" 149,500,000 even though their difference is 500,000. The thing which makesone number "close to" another is that their ratio should be close to 1, not their difference close to 0. It's a natural way to
compare, just as we're doing when we compare p(x) to q(x) via the ratio p(x)q(x)
. The ratio .00002.00001
is 2 (not very close to 1)
whereas 150,000,000149,500,000
= 1.003 so the latter numbers are close to one another ... in this special sense of the phrase "close to".
S: Okay, so when you say cos x is "close to" 1 - x2
2 then you're really saying that their ratio is close to 1, right?
P: Right.
S: Hah! Gotcha! Their difference is also close to zero! See? cos x - (1 - x2
2 ) has a limit of 0 as x->0, so their difference is close
to 0!
P: That's just an accident. I mean, it doesn't always happen that way. Sometimes p(x)q(x)
->1 yet p(x) - q(x) doesn't ->0. See? One
has a tendency to say pq ≈ 1 hence p ≈ q hence p - q ≈ 0 and that could be wrong!
S: Prove it.
P: Let's see ... uh, yes ... let's try it with p(x) = 1
x2 +
1x and q(x) =
1
x2 . Then
pq =
1
x2 +
1x
1
x2
= 1+x ->1 as x->0 hence p is "close
to" q when x is small (in our special meaning of the phrase "close to") so we might think that p(x) - q(x) ->0, but we'd be
wrong since p(x) - q(x) = 1x which certainly doesn't approach zero as x->0! It's like the sun and the asteroid; because their
ratio is very close to 1 doesn't mean their difference is small. Got it?S: Got it. However, before you go on, tell me why you said, earlier, that ... to use your exact words:
limx->0
1 - cos x
x2 =
12 tells us that 1 - cos x ≈
x2
2 hence cos x ≈ 1 -
x2
2 when x is small.
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You just said you couldn't do this, didn't you?P: You're right. Mia culpa. I apologize ... BUT the statement is still true, even though the reasoning is fallacious.S: Can I do that on an exam? I mean, get the right result with the wrong reasoning?P: Sure ... but you won't get any marks for it. But hold on, let me do it properly so you can see why the final statement is true.
First, cos x ≈ 1 - x2
2 means that cos x -
1 - x2
2 -> 0 (as x->0), so that's what we must prove, starting with the known result
limx->0
1 - cos x
x2 =
12 which we got using l'Hopital's rule. Now
limx->0
1 - cos x
x2 =
12 means (according to the definition of
"limit") that, for any choice of error e we can make
1 - cos x
x2 - 12 < e by restricting x to lie in some small interval about
x = 0, say 0 < | x | < h. Hence - e < 1 - cos x
x2 -
12 < e and we can reorganize this to read:
- ex2 < cos x -
1 - x2
2 < ex2. Now let x->0 and get (using the ol' SQUEEZE theorem):
limx->0
(- ex2) ≤
limx->0
[ ]cos x - (1 - x2/2) ≤
limx->0
(ex2) and since the outside limits are 0, then
limx->0
[ ]cos x - (1 - x2/2) = 0 as well.
S: Do I have to know this for the final exam?P: No. I just thought you might be interested.S: Wrong!
Example: Evaluate
limx->0
1
sin x - 1x .
Solution: We must reorganize the expression 1
sin x - 1x (which now has the form ∞ - ∞ as x->0) , so it has
the form 00 , so we rewrite it as
x - sin xx sin x . Using l'Hopital's rule twice we get:
limx->0
x - sin xx sin x =
limx->0
1 - cos x
x cos x + sin x =
limx->0
sin x
2 cos x - x sin x = 02 = 0.
Note: this says that not only is sin x ≈ x but also 1
sin x ≈ 1x (which may be a little surprising) .
An interesting question: If p(x) - q(x) -> 0, does 1
p(x) - 1
q(x) -> 0? Answer? Sometimes, but not always. See if
you can find a simple example where, say,
limx->0
(p(x) - q(x)) = 0, yet
limx->0
1
p(x) - 1
q(x) ≠ 0
LECTURE 13
POLAR COORDINATES
We usually describe the location of points on a plane by giving the distance left-right and distance up-downfrom some origin. Sometimes we think of it as distance east-west and north-south. Perhaps we think of latitudeand longitude. In any case, they are "rectangular" or "Cartesian" coordinates for the point ... but they aren't theonly way to describe the location of a point and, indeed, sometimes rectangular coordinates are a terrible choice.
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For example, if you're in a desert and were describing how to getsomewhere you wouldn't say "go 3 km east then 4 km north"(using rectangular coordinates). You're more likely to saysomething like "Go 5 miles north-east", giving a distance and adirection. These are POLAR COORDINATES. The distancefrom the origin is called r and the direction is specified by theangle q measured (in RADIANS!) counterclockwise from thepositive x-direction as shown ===>>>It's clear from the diagram that there is a simple relationshipbetween the rectangular coordinates of the point P, namely (x,y),and the polar coordinates (r,q). These are:
x = r cos q, y = r sin q and r2 = x2 + y2, tan q = yx
Before we continue with polar coordinates, let's digress toconsider yet another coordinate system (just so you don't thinkthese are the only ones). Whereas (x,y) provides 2 distances and(r,q) gives a distance and an angle, we might consider acoordinate system which describes the location of a point in theplane using two angles. We pick two origins (why not?) and givetwo angles q1 and q2 as shown.
There are major problems with this (q1, q2) coordinate system. If
q1 > q2 , then the two rays emanating from the origins don't
intersect anywhere. Further, what are the coordinates of thepoints P1, P2 and P3 shown? They are P1(π,π) and P2(0,π) and
P3(0,0) but they aren't the only points with these coordinates! In fact every point between the two origins has the
same coordinates, namely (0,π). Because of this we won't spend any more time on this coordinate system!!
Back to polar coordinates:In identifying a point (r,q) it is convenient to think of "polar"graph paper. Note that "rectangular" graph paper has the curves x= constant and y = constant drawn for you (such as x = -2, x = -1, x = 0 etc.) whereas polar graph paper has the curves r= constant and q = constant drawn ==>>Until you are accustomed to thinking in polar coordinates, it isconvenient to convert polar equations (such as r = 2 or q = π/4)
into rectangular coordinates: r = 2 becomes x2 + y2 = 22, a circleof radius 2 and centre the origin. q = π/4 becomes tan q = 1 = y/xso y = x (a line through the origin with slope 1).The diagram illustrates various points ===>If the circles shown are described by r = 1, r = 2, etc. then p1 has
polar coordinates (4,π/4) and p2(3,π/6), p3(4,5π/6), p4(4,5π/4),
p5(3,4π/3) and p6(2,7π/4). However, every point in the plane has
a whole host of polar coordinates! The point p6 can also be
described by r = 2 and q = - π/4 (with negative q correspondingto clockwise measurements of the angle). We also havep1(4,9π/4) where the angle q corresponds to a complete
revolution of 2π plus an additional π/4. Indeed, p1 can be
described by r = 2 and q = 2πn + π/4 for any integer "n". Butthat's not all. If we regard negative q as meaning "move oppositeto the positive direction" then we can also consider negative
values for r, also meaning "move opposite to the positive direction". The positive direction is in the direction of q soa negative r means move in the direction opposite to this. That gives to p1 (for example) the polar coordinates
r = - 4 and q = π/4 + π = 5π/4. In spite of the plethora of polar coordinates for each point in the plane, we usuallyidentify a point with a positive r and an angle (in RADIANS!) in the interval 0 ≤ q < 2π.
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Now for some more interesting polar curves:Whereas the equation y = x (in rectangular coordinates) describes a straight line through the origin with slope 1,r = q describes quite a different curve in polar coordinates. As the q increases, r increases and the points withpolar coordinates satisfying r = q lie on a spiral. Although the computer-plotted graph below shows r = q forpositive values of q (hence of r), we can also plot r = q for negative q as well. See what it'll look like? Justreplace the polar coordinates of every point by their negatives. For example, (-2π, -2π) and (-π/2, -π/2) alsosatisfy r = q.
Consider the polar curve r = 2 cos q. We'll convert to rectangular coordinates, hoping to recognize the curve. To
do this we want to create terms r2 and/or r cos q and/or r sin q and/or tan q which we'll replace by x2+y2, x, y
and y/x respectively. Multiplying by r gives r2 = 2 r cos q hence x2+y2 = 2x is the equivalent rectangular
equation and this can be rewritten (x - 1)2 + y2 = 1, a circle of radius "1" with centre at (1,0).After a while it gets tiring having to convert to rectangular coordinates ... and often we don't recognize therectangular equation anyway, so we should get accustomed to sketching directly in polar coordinates.In order to sketch a polar curve r = f(q) it's convenient to first sketch this relation as though r and q wererectangular coordinates (or we could make a table of values, but a picture is worth a thousand tables).
In the above example, we've identified 4 points at q = 0, q = a (where r is a local maximum), q = b (where r is alocal minimum) and q = c ... then we show these points in polar coordinates, noting where r is decreasing andincreasing.But something wonderful happens if f(q) is negative, as shown in the example below:
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Again we sketch r = f(q) and identify a few points of interest at q = 0, a, b, c, d and e. Note that, whenq = b, r has decreased to zero (so the distance from the origin, in polar coordinates, is zero and the polar curve goes
through the origin). From q = b to q = d, r = f(q) is negative and although the q-direction is (roughly) into thesecond quadrant, the negative r-values indicate that we move opposite to that direction, placing the points(roughly) in the fourth quadrant. In particular, at q = c (which points roughly north-west, in polar coordinates), rhas its most negative value ... so we move south-east.
Examples: The polar curves, r = 1
1+a cos q are conic sections (hyperbolas, parabolas, circles, ellipses),
depending upon the value of "a". For each of the following a-values, plot the curve and identify:
(a) a = 0 (b) a = .5 (c) a = 1 (d) a = 2Solutions:(a) a = 0 is easiest. The polar equation is r = 1, a circle.
(b) For a = .5, r is always positive and goes from a minimum value of 1
1+.5 = 23 (when q = 0) to a
maximum of 1
1-.5 = 2 (when q = π). It's the ellipse.
(c) For a = 1, r = 1
1+cos q has its minimum of 12 at q = 0 but increases to ∞ as q->π (the denominator becoming
zero). Or, to put it differently,
limq->π
r = ∞ and q = π gives the only infinity. This is the parabola.
(d) For a = 2, r = 1
1+2 cos q not only becomes infinite but does so twice ... and r is also negative.
From q=0 to q=2π3 , r increases from its initial value of
13 , becoming infinite as q->
2π3 (since 1+2 cos q->+0) .
From q=2π3 to q=
4π3 , r <0 (since 1+2 cos q<0) and the polar curve lies in the fourth, then the first quadrant.
From q=4π3 to q=2π, r is again positive and returns from infinity (when q =
4π3 ) to
13 (at q = 2π) .
The curve is a hyperbola and it has two branches, the right-most branch being traversed while r is negative and qpoints into the second and third quadrants.
S: That's hard, isn't it? I mean, do you really expect ...
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P: Don't worry. I only hope you can follow the arguments I gave, and could sketch these curves if you had enough time, but Iwon't expect you to reproduce them on an exam.
S: Whew!P: But they are nice, aren't they? In fact, these polar equations describe the conic sections with a focus at the origin and that's
different than the usual rectangular equations. In fact, if you wanted to describe the orbit of heavenly objects as they movedabout the sun then you'd likely pick the sun as focus and you'd get one of these polar equations for the orbit of planets orcomets. Nice, eh?
S: Wonderful ...
Examples: Sketch r2 = sin 2qSolution: Note that sin 2q is negative when π < 2q < 2π (since that puts 2q in the third or fourth quadrant
where the sine function is negative) ... and again when 3π < 2q < 4π. Hence, when π2 < q < π and again when
3π2 < q < 2π, there is no curve!! That's because r2 = sin 2q cannot be negative. For q-values in between (starting at q
= 0), r2 increases to a maximum when 2q = π2 (i.e. q =
π4 ) then decreases to zero when 2q = π (i.e. q =
π2 ).
Then we come to the sector where there is no curve, then we start again with q = π where r increases to a
maximum of 1 when 2q = 5π2 (i.e. q =
5π4 ) and decreases again to zero for 2q = 4π, or q = 2π ... and that takes
us through one complete circuit of the origin and if we continue, the curve repeats.
This polar curve is called a LEMNISCATE*.
S: You say r2 can't be negative ... so there's no curve there. That's weird, isn't it?
P: You've seen it before, you just don't recognize it. In fact, if you sketch y2 = 4 - x2 in rectangular coordinates you'd say that,
for x2 > 4, there is no curve because y2 can't be negative ... hence the curve lies only in x2≤ 4.
S: I'd never say that! I don't even recognize y2 = 4 - x2!
P: How about x2 + y2 = 4?
* This is often called the lemniscate of Bernoulli. The Bernoulli family included Jakob (1654-1705) who taughthimself the Newton/Leibniz calculus and invented polar coordinates, and his younger brother Johann (1667-1748)who studied medicine as well as calculus, and Johann's son, Daniel (1700-1782) who became an outstandingmathematical physicist. There was great rivalry among the Bernoullis. As the story goes, Johann posed a problemto the mathematicians of the world, the Brachistochrone Problem: "A wire is bent into a curve joining two givenpoints. A bead slides down the wire without friction. What curve will give the minimum time of descent?" Theproblem was solved by Newton, Leibniz (the curve is a CYCLOID) ... and Jakob. Johann was not pleased.
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S: Aah, that's different. Anyway, why didn't you sketch r2 = sin 2q in rectangular coordinates first, as you suggested we do?P: Okay, I'll do it that way. First I sketch r = sin 2q (which is easy) then I take ± its square root to get the graph of
r2 = sin 2q (but only where sin 2q is positive, of course).
S: How's that again!?
P: It's how you can sketch y2 = f(x). Just sketch y = f(x), then throw away all the negative pieces of f(x) (because there's no
curve there, remember?), then take ± f(x) with what's left. That's y2 = f(x).
Then you get those cute little loops wherever f(x) is positive.The method is something like sketching y = | f(x) |. We first sketch y = f(x), then reflect all the negative pieces of f(x) in thex-axis, i.e. replace them with - f(x). Remember?
S: But the curve is different, right? I mean, y = f(x) doesn't look like y = f(x), does it?
P: No, but when f(x) increases or decreases, so does f(x) and that's enough to sketch y = f(x) ... uh, except for one otherthing which I almost hate to mention.
S: Go ahead. Nothing'll scare me now.
P: Well, where f(x) is zero, the graph of y = f(x) usually has a vertical tangent (i.e the derivative is infinite). That's because,
for y = f(x) , we have dydx
= ddx
( )f(x) 1/2= 12 ( )f(x) -1/2 f ' (x) =
f ' (x)
2 f(x) which becomes infinite as f(x)->0 (unless, of
course f ' (x)->0 as well ... in which case we've got a 00 form and
dydx
needn't be infinite).
S: That does scare me.
Examples: Sketch each of the following:(a) r = 4 + 3 cos q (b) r = 4 + 4 cos q(c) r = 4 + 5 cos q
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Solutions:All are plotted on the same graph. Note that r =4 + 3 cos q has max and min r-values of 7 and1 and they occur at q = 0 and q = π(respectively) while r = 4 + 4 cos q has maxand min r-values of 8 and 0, the latter being atthe origin (and occurs for q = π). Finally, thelast curve has negative r-values which occurwhenever
4 + 5 cos q < 0, hence when q lies between twoparticular angles in the second and thirdquadrant. For q in this interval, although the q-direction is west of the y-axis, the negative r-value places the point east of the y-axis ...generating a small loop with right-most point r= -1, q = π.All are called LIMAÇONS (with the form r = a+ b cos q) although the middle curve, where b= a, is more commonly known as aCARDIOID. When b > a there's an inner loop.
INTERSECTION OF POLAR CURVES:A problem we'll meet shortly involves finding the points of intersections of two polar curves, say r = f(q) andr = g(q). The procedure is much the same as for rectangular coordinates: solve these 2 equations in 2 unknowns:r and q. Because there are an infinite variety of ways to describe a single point in polar coordinates (i.e. (r,q) and(r,q+2π) and (-r,q+π) etc. etc.) it's best to sketch the polar curve and anticipate the location of points ofintersection. That way you'll know when you have them all!
Example: Find the point(s) of intersection of r = cos q and r = 1 - cos q.Solution: A sketch indicates two points of intersection: one in the first
and one in the fourth quadrant. To find them, set cos q = 1 - cos q so cos q
= 12 hence q =
π3 (that's the one in the first quadrant) and q =
5π3 or,
perhaps this one's simpler to describe as q = - π3 . In any case r = cos
π3 =
12 so the
two points are (12 , ±
π3 ). Note that r = 1 - cos q is a cardioid with its minimum r-value (namely r = 0) occurring at q
= 0 and its maximum (r = 2) at q = π. Also, r = cos q is a circle as we've already seen.
PS:S: I haven't seen ... have I?P: Well, I didn't sketch it last time. Pay attention: r = cos q has its maximum (r = 1) at q = 0, then r decreases as q increases
until, when q = π2 , we have r = 0 (so the curve is now at the origin) and after that, q points into the second quadrant but now
r is negative so we move opposite to the q-direction and get the curve going into the fourth quadrant (the lower half of thecircle) until q = π and r = -1 which actually gives the right-most point on this circle even though q points due west (becauser is negative, so the curve is east) and now q continues into the third quadrant but r is still negative so the curve is traced outin the first quadrant, again(!), retracing the upper half of the circle, then finally q points into the fourth quadrant and rbecomes positive again and we go in the q-direction and retrace the lower half of the circle (again!) so we stand back andnotice that when q goes from 0 to 2π the curve r = cos q is actually traced out twice! Got it?
S: zzzzz
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P: Wake up ... here's one for you. Sketch the FOLIUM:
(x2+y2)2 = x2 y, but I'd suggest changing first to polars. And whileyou're at it, tell me the maximum value of r (i.e the maximum distancefrom the origin). And while you're at it, notice how many times thecurve is traced when q goes from 0 to 2π. Got it? Okay, I'll do it myself.
Put x2 + y2= r2 and x = r cos q and y = r sin q and get
r4 = r3 cos2q sin q so we cancel r3 from each side, checking to see if
we've thrown any points away and noticing that the origin has r = 0 but we're not throwing this point away because it's
satisfies the equation that's left, namely r = cos2q sin q (in fact we get to the origin every time cos q = 0 or sin q = 0) andnow we imagine q increasing from q = 0 (where we begin, at the origin) so r increases to some maximum value (and we'llfind out where that occurs in a minute) then r decreases again to 0 when q = π/2 ('cause cos q = 0) then q points into the
second quadrant and cos q goes negative but r is still positive 'cause we've got a cos2q in the equation, but soon we get to q= π where sin q = 0 so we go to the origin again and for the rest of the time q points into the third and fourth quadrants but ris negative (because of the sin q) so the curve is traced out in the first and second quadrants (again!) until, finally, q = 2πand we re-arrive where we began, at the origin, ready to retrace the curve for a third time should q decide to continue. Gotit?
S: You forgot to find the maximum r-value.
P: Aah yes, well, we must maximize the continuous function r = cos2q sin q on the closed interval 0 ≤ q ≤ π ... got that? acontinuous function on a closed interval ... so we find the critical points within this interval ... wheredrdq
= cos2q (cos q) + 2 cos q (-sin q) sin q = 0 and that means that cos3q - 2 cos q sin2q = 0 or, let's see, I can factor this so
it reads: cos q (cos2q - 2 sin2q) = 0 hence either cos q = 0 (meaning q = π2 ) or cos2q - 2 sin2q = 0 which I can rewrite,
putting cos2q = 1 - sin2q, and I get 1 - 3 sin2q = 0 so sin q = 1
3 and there are two q-values in 0 < q < π which have this
sine and each yields a maximum for r (one's in the first and one in the second quadrant, at the points labelled p and q). See?S: You still haven't found the maximum r-value ... and you have to check the end-points to find the maximum ... and you
forgot that 1 - 3 sin2q = 0 means sin q = ±1
3 , so you forgot that too.
P: Okay, the end-points ... q = 0 and q = π each give r = 0 (certainly not the maximum). Also, I only need to consider q's in
the interval 0 < q < π, where sin q > 0 (so I ignore sin q = - 1/ 3 ), but I guess I should find the maximum r-value. I
substitute the appropriate q into r = cos2q sin q = (1 - sin2q) sin q and get rmax = (1 - 13 )
1
3 =
2
3 3 .
S: You can't leave it in that form ... rationalize the denominator. I learned that in kindergarten.P: Go back to sleep.